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PutnamBench

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putnam_1997_b2

theorem putnam_1997_b2 (f g : ℝ → ℝ) (hg : ∀ x : ℝ, g x ≥ 0) (hfderiv1 : ContDiff ℝ 1 f) (hfderiv2 : Differentiable ℝ (deriv f)) (hfg : ∀ x : ℝ, f x + iteratedDeriv 2 f x = -x * g x * deriv f x) : IsBounded (range (fun x => |f x|)) := sorry

Let $f$ be a twice-differentiable real-valued function satisfying \[f(x)+f''(x)=-xg(x)f'(x),\] where $g(x)\geq 0$ for all real $x$. Prove that $|f(x)|$ is bounded.

['analysis']

Section putnam_1997_b2. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Theorem putnam_1997_b2: forall (f g: R -> R), exists (M: R), forall (x: R), ex_derive_n f 2 x /\ g x >= 0 /\ f x + Derive_n f 2 x = -x * g x * Derive f x -> -M <= f x <= M. Proof. Admitted. End putnam_1997_b2.

theory putnam_1997_b2 imports Complex_Main "HOL-Analysis.Derivative" begin theorem putnam_1997_b2: fixes f g :: "real \<Rightarrow> real" assumes hg : "\<forall> x :: real. g x \<ge> 0" and hfdiff : "(f differentiable_on UNIV) \<and> (deriv f) differentiable_on UNIV" and hfg : "\<forall> x :: real. f x + (deriv^^2) f x = -x * g x * (deriv f x)" shows "bounded (image f UNIV)" sorry end

putnam_1997_b3

abbrev putnam_1997_b3_solution : Set ℕ := sorry -- {n | (1 ≤ n ∧ n ≤ 4) ∨ (20 ≤ n ∧ n ≤ 24) ∨ (100 ≤ n ∧ n ≤ 104) ∨ (120 ≤ n ∧ n ≤ 124)} theorem putnam_1997_b3 (n : ℕ) (hn : n > 0) : n ∈ putnam_1997_b3_solution ↔ ¬5 ∣ (∑ m in Finset.Icc 1 n, 1/m : ℚ).den := sorry

For each positive integer $n$, write the sum $\sum_{m=1}^n 1/m$ in the form $p_n/q_n$, where $p_n$ and $q_n$ are relatively prime positive integers. Determine all $n$ such that 5 does not divide $q_n$.

Show that the solution is the set of natural numbers which are between $1$ and $4$, or between $20$ and $24$, or between $100$ and $104$, or between $120$ and $124$.

['number_theory']

Section putnam_1997_b3. Require Import Reals Coquelicot.Coquelicot. From mathcomp Require Import div. Open Scope R. Definition putnam_1997_b3_solution (n: nat) := and (le 1 n) (le n 4) \/ and (le 20 n) (le n 24) \/ and (le 100 n) (le n 104) \/ and (le 120 n) (le n 124). Theorem putnam_1997_b3: forall (n: nat), gt n 0 -> exists (p q: nat), gt p 0 /\ gt q 0 /\ coprime p q = true /\ sum_n (fun m => 1 / INR (m + 1)) n = INR p / INR q -> neq (q mod 5) 0 -> putnam_1997_b3_solution n. Proof. Admitted. End putnam_1997_b3.

theory putnam_1997_b3 imports Complex_Main begin definition putnam_1997_b3_solution :: "nat set" where "putnam_1997_b3_solution \<equiv> undefined" (* {n :: nat. (1 \<le> n \<and> n \<le> 4) \<or> (20 \<le> n \<and> n \<le> 24) \<or> (100 \<le> n \<and> n \<le> 104) \<or> (120 \<le> n \<and> n \<le> 124)} *) theorem putnam_1997_b3: fixes n :: "nat" assumes hn : "n > 0" shows "n \<in> putnam_1997_b3_solution \<longleftrightarrow> \<not> (5 dvd (snd (quotient_of (\<Sum> m \<in> {1..n}. 1/(rat_of_nat m)))))" sorry end

putnam_1997_b4

theorem putnam_1997_b4 (a : ℕ → ℕ → ℤ) (ha : ∀ n m : ℕ, a n m = coeff ((1 + X + X^2)^m) n) : ∀ k : ℕ, k ≥ 0 → (∑ i in Finset.Icc 0 (Nat.floor (2*k/(3 : ℚ))), (-1)^(i : ℕ) * (a (k - i) i : ℝ)) ∈ Set.Icc (0 : ℝ) 1 := sorry

Let $a_{m,n}$ denote the coefficient of $x^n$ in the expansion of $(1+x+x^2)^m$. Prove that for all [integers] $k\geq 0$, \[0\leq \sum_{i=0}^{\lfloor \frac{2k}{3}\rfloor} (-1)^i a_{k-i,i}\leq 1.\]

['algebra']

theory putnam_1997_b4 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_1997_b4: fixes a :: "nat \<Rightarrow> nat \<Rightarrow> int" assumes ha : "\<forall> n m :: nat. a n m = coeff ((monom 1 0 + monom 1 1 + monom 1 2)^m) n" shows "\<forall> k :: nat \<ge> 0. (\<Sum> i \<in> {0..nat \<lfloor>((2*(rat_of_nat k)))/(rat_of_nat 3)\<rfloor>}. (-1)^(i) * (a (k-i) i)) \<in> {0 :: real..1}" sorry end

putnam_1997_b5

theorem putnam_1997_b5 (n : ℕ) (hn : n ≥ 2) : tetration 2 n ≡ tetration 2 (n-1) [MOD n] := sorry

Prove that for $n\geq 2$, \[\overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n$ terms}} \equiv \overbrace{2^{2^{\cdots^{2}}}}^{\mbox{$n-1$ terms}} \quad \pmod{n}.\]

['number_theory']

Section putnam_1997_b5. Require Import Nat. Theorem putnam_1997_b5: let fix pow_n (b n: nat) : nat := match n with | O => 1 | S n' => b * pow_n b n' end in forall (n: nat), n >= 2 -> pow_n 2 n-1 mod n = pow_n 2 n-2. Proof. Admitted. End putnam_1997_b5.

theory putnam_1997_b5 imports Complex_Main "HOL-Number_Theory.Cong" begin theorem putnam_1997_b5: fixes n :: "nat" and tetration :: "nat \<Rightarrow> nat \<Rightarrow> nat" assumes hn : "n \<ge> 2" and htetrationbase : "\<forall> a :: nat. tetration a 0 = 1" and htetration : "\<forall> a n :: nat. tetration a (n+1) = a^(tetration a n)" shows "[tetration 2 n = tetration 2 (n-1)] (mod n)" sorry end

putnam_2005_a1

theorem putnam_2005_a1 : ∀ n : ℤ, n > 0 → (∃ k : ℕ, ∃ a : Fin k → Fin 2 → ℕ, n = ∑ i : Fin k, 2^(a i 0)*3^(a i 1) ∧ (∀ i j : Fin k, i ≠ j → ¬(2^(a i 0)*3^(a i 0) ∣ 2^(a j 0)*3^(a j 1)))) := sorry

Show that every positive integer is a sum of one or more numbers of the form $2^r 3^s$, where $r$ and $s$ are nonnegative integers and no summand divides another.

['number_theory']

Section putnam_2005_a1. Require Import Nat List Coquelicot.Coquelicot. Theorem putnam_2005_a1: forall (n: nat), n > 0 -> exists (l: list nat), forall (p q: nat), In p l /\ In q l -> exists (r1 s1 r2 s2: nat), p = 2 ^ r1 * 3 ^ s1 /\ q = 2 ^ r2 * 3 ^ s2 /\ p mod q <> 0 /\ q mod p <> 0. Proof. Admitted. End putnam_2005_a1.

theory putnam_2005_a1 imports Complex_Main begin theorem putnam_2005_a1: shows "\<forall> n :: int. n > 0 \<longrightarrow> (\<exists> k :: nat. \<exists> a :: nat \<Rightarrow> (nat \<times> nat). n = (\<Sum> i\<in>{1..k}. 2^(fst (a i)) * 3^(snd (a i))) \<and> (\<forall> i \<in> {1..k}. \<forall> j \<in> {1..k}. i \<noteq> j \<longrightarrow> (\<not> (2^(fst (a i)) * 3^(snd (a i)) dvd 2^(fst (a j)) * 3^(snd (a j))))))" sorry end

putnam_2005_a3

theorem putnam_2005_a3 (p : Polynomial ℂ) (n : ℕ) (g : ℂ → ℂ) (pdeg : p.degree = n) (pzeros : ∀ z : ℂ, p.eval z = 0 → Complex.abs z = 1) (hg : ∀ z : ℂ, g z = (p.eval z) / z ^ ((n : ℂ) / 2)) : ∀ z : ℂ, (deriv g z = 0) → (Complex.abs z = 1) := sorry

Let $p(z)$ be a polynomial of degree $n$ all of whose zeros have absolute value $1$ in the complex plane. Put $g(z)=p(z)/z^{n/2}$. Show that all zeros of $g'(z)=0$ have absolute value $1$.

['analysis', 'algebra']

Section putnam_2005_a3. Require Import Reals Coquelicot.Coquelicot. From Coqtail Require Import Cpow. Theorem putnam_2005_a3: forall (c: nat -> Z) (n: nat), let p (z: C) := sum_n (fun i => IZR (c i) * Cpow z i) (n + 1) in forall (r: C), p r = 0 -> r = RtoC (-1) /\ r = RtoC 1 -> let g (z: C) := p z / Cpow z (n / 2) in forall (r: C), g r = 0 -> r = RtoC (-1) /\ r = RtoC 1. Proof. Admitted. End putnam_2005_a3.

theory putnam_2005_a3 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2005_a3: fixes p :: "complex poly" and n :: nat and g :: "complex \<Rightarrow> complex" assumes pdeg: "degree p = n" and pzeros: "\<forall>z::complex. (poly p z = 0 \<longrightarrow> norm z = 1)" and hg: "\<forall>z::complex. g z = (poly p z) / csqrt (z^n)" shows "\<forall>z::complex. (deriv g z = 0 \<longrightarrow> norm z = 1)" sorry end

putnam_2005_a4

theorem putnam_2005_a4 (n : ℕ) (H : Matrix (Fin n) (Fin n) ℝ) (a b : ℕ) (S : Matrix (Fin a) (Fin b) ℝ) (npos : n ≥ 1) (Hentries : ∀ i j : Fin n, H i j = 1 ∨ H i j = -1) (Hortho : H.HasOrthogonalRows) (hab : 1 ≤ a ∧ a ≤ n ∧ 1 ≤ b ∧ b ≤ n) (Ssub : ∃ (rri : Fin a → Fin n) (cri : Fin b → Fin n), rri.Injective ∧ cri.Injective ∧ S = H.submatrix rri cri) (Sentries : ∀ (i : Fin a) (j : Fin b), S i j = 1) : a * b ≤ n := sorry

Let $H$ be an $n \times n$ matrix all of whose entries are $\pm 1$ and whose rows are mutually orthogonal. Suppose $H$ has an $a \times b$ submatrix whose entries are all $1$. Show that $ab \leq n$.

['linear_algebra']

theory putnam_2005_a4 imports Complex_Main "HOL-Analysis.Derivative" begin theorem putnam_2005_a4: fixes n :: nat and H :: "real^'n^'n" and a b :: nat and S :: "real^'b^'a" assumes npos: "n \<ge> 1" and ncard: "CARD('n) = n" and Hentries: "\<forall>i j::'n. H$i$j = 1 \<or> H$i$j = -1" and Hortho: "\<forall>i j::'n. (i \<noteq> j \<longrightarrow> orthogonal (row i H) (row j H))" and hab: "1 \<le> a \<and> a \<le> n \<and> 1 \<le> b \<and> b \<le> n" and abcard: "CARD('a) = a \<and> CARD('b) = b" and Ssub: "\<exists>(rri::'a\<Rightarrow>'n)(cri::'b\<Rightarrow>'n). inj rri \<and> inj cri \<and> (\<forall>(i::'a)(j::'b). S$i$j = H$(rri i)$(cri j))" and Sentries: "\<forall>(i::'a)(j::'b). S$i$j = 1" shows "a*b \<le> n" sorry end

putnam_2005_a5

abbrev putnam_2005_a5_solution : ℝ := sorry -- Real.pi * (Real.log 2) / 8 theorem putnam_2005_a5 : ∫ x in (0:ℝ)..1, (Real.log (x+1))/(x^2 + 1) = putnam_2005_a5_solution := sorry

Evaluate $\int_0^1 \frac{\ln(x+1)}{x^2+1}\,dx$.

Show that the solution is $\pi / 8 * \log 2$.

['analysis']

Section putnam_2005_a5. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2005_a5_solution := PI * (ln 2 / ln 10) / 8. Theorem putnam_2005_a5: RInt (fun x => ln (x + 1) / (x ^ 2 + 1)) 0 1 = putnam_2005_a5_solution. Proof. Admitted. End putnam_2005_a5.

theory putnam_2005_a5 imports Complex_Main "HOL-Analysis.Lebesgue_Measure" "HOL-Analysis.Set_Integral" begin definition putnam_2005_a5_solution :: "real" where "putnam_2005_a5_solution \<equiv> undefined" (* pi * (ln 2) / 8 *) theorem putnam_2005_a5: shows "interval_lebesgue_integral lebesgue 0 1 (\<lambda> x :: real. (ln (x + 1))/(x^2 + 1)) = putnam_2005_a5_solution" sorry end

putnam_2005_b1

abbrev putnam_2005_b1_solution : MvPolynomial (Fin 2) ℝ := sorry -- (MvPolynomial.X 1 - 2 * MvPolynomial.X 0) * (MvPolynomial.X 1 - 2 * MvPolynomial.X 0 - 1) theorem putnam_2005_b1 : putnam_2005_b1_solution ≠ 0 ∧ ∀ a : ℝ, MvPolynomial.eval (fun n : Fin 2 => if (n = 0) then (Int.floor a : ℝ) else (Int.floor (2 * a))) putnam_2005_b1_solution = 0 := sorry

Find a nonzero polynomial $P(x,y)$ such that $P(\lfloor a \rfloor,\lfloor 2a \rfloor)=0$ for all real numbers $a$. (Note: $\lfloor \nu \rfloor$ is the greatest integer less than or equal to $\nu$.)

Show that $P(x,y)=(y-2x)(y-2x-1)$ works.

['algebra']

Section putnam_2005_b1. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2005_b1_solution (c: nat -> nat -> R) (n m: nat) := n = 2%nat /\ m = 2%nat /\ c = fun x y => match x, y with | 0, 1 => -1 | 0, 2 => 1 | 1, 0 => 2 | 1, 1 => -4 | 2, 0 => 4 | _, _ => 0 end. Theorem putnam_2005_b1: exists (c: nat -> nat -> R) (n m: nat), let p (x y: R) := sum_n (fun i => (sum_n (fun j => c i j * x ^ i * y ^ j) m)) (n + 1) in forall (a: R), p (IZR (floor a)) (IZR (floor (2 * a))) = 0 <-> putnam_2005_b1_solution c n m. Proof. Admitted. End putnam_2005_b1.

theory putnam_2005_b1 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin (* there might be multiple possible correct answers *) definition putnam_2005_b1_solution :: "real poly poly" where "putnam_2005_b1_solution \<equiv> undefined" (* (monom (monom 1 0) 1 - 2*(monom (monom 1 1) 0)) * (monom (monom 1 0) 1 - 2*(monom (monom 1 1) 0) - 1) *) theorem putnam_2005_b1: shows "putnam_2005_b1_solution \<noteq> 0 \<and> (\<forall>a::real. poly (poly putnam_2005_b1_solution [:\<lfloor>2*a\<rfloor>:]) \<lfloor>a\<rfloor> = 0)" sorry end

putnam_2005_b2

abbrev putnam_2005_b2_solution : Set (ℕ × (ℕ → ℤ)) := sorry -- {(n, k) : ℕ × (ℕ → ℤ) | (n = 1 ∧ k 0 = 1) ∨ (n = 3 ∧ (k '' {0, 1, 2} = {2, 3, 6})) ∨ (n = 4 ∧ (∀ i : Fin 4, k i = 4))} theorem putnam_2005_b2 (n : ℕ) (k : ℕ → ℤ) (nkpos : Prop) (nkeq1 : Prop) (nkeq2 : Prop) (hnkpos : nkpos = (n > 0 ∧ (∀ i : Fin n, k i > 0))) (hnkeq1 : nkeq1 = (∑ i : Fin n, k i = 5 * n - 4)) (hnkeq2 : nkeq2 = (∑ i : Fin n, (1 : ℝ) / (k i) = 1)) : (nkpos ∧ nkeq1 ∧ nkeq2) ↔ (n, k) ∈ putnam_2005_b2_solution := sorry

Find all positive integers $n,k_1,\dots,k_n$ such that $k_1+\cdots+k_n=5n-4$ and $\frac{1}{k_1}+\cdots+\frac{1}{k_n}=1$.

Show that the solutions are $n=1$ and $k_1=1$, $n=3$ and $(k_1,k_2,k_3)$ is a permutation of $(2,3,6)$, and $n=4$ and $(k_1,k_2,k_3,k_4)=(4,4,4,4)$.

['algebra']

Section putnam_2005_b2. Require Import Nat List Reals Coquelicot.Coquelicot. Import ListNotations. Definition putnam_2005_b2_solution (n: nat) (k: list nat) := (n, k) = (1%nat, [1%nat]) \/ (n, k) = (3%nat, [2%nat; 3%nat; 6%nat]) \/ (n, k) = (3%nat, [2%nat; 6%nat; 3%nat]) \/ (n, k) = (3%nat, [3%nat; 2%nat; 6%nat]) \/ (n, k) = (3%nat, [3%nat; 6%nat; 2%nat]) \/ (n, k) = (3%nat, [6%nat; 2%nat; 3%nat]) \/ (n, k) = (3%nat, [6%nat; 3%nat; 2%nat]) \/ (n, k) = (4%nat, [4%nat; 4%nat; 4%nat; 4%nat]). Theorem putnam_2005_b2: forall (n: nat) (k: list nat), forall (x: nat), fold_left add k 0%nat = sub (mul 5 n) 4 /\ sum_n (fun n => 1 / INR (nth n k 0%nat)) n = 1 <-> putnam_2005_b2_solution n k. Proof. Admitted. End putnam_2005_b2.

theory putnam_2005_b2 imports Complex_Main begin (* uses (nat \<Rightarrow> nat) instead of (Fin n \<Rightarrow> nat) *) definition putnam_2005_b2_solution :: "(nat \<times> (nat \<Rightarrow> nat)) set" where "putnam_2005_b2_solution \<equiv> undefined" (* {(n::nat,k::nat\<Rightarrow>nat). (n = 1 \<and> k 0 = 1) \<or> (n = 3 \<and> k ` {0,1,2} = {2,3,6}) \<or> (n = 4 \<and> (\<forall>i::nat\<in>{0..3}. k i = 4))} *) theorem putnam_2005_b2: fixes n :: nat and k :: "nat \<Rightarrow> nat" and nkpos :: bool and nkeq1 :: bool and nkeq2 :: bool assumes "nkpos \<equiv> (n > 0 \<and> (\<forall>i::nat\<in>{0..(n-1)}. k i > 0))" and "nkeq1 \<equiv> ((\<Sum>i::nat=0..(n-1). k i) = 5*n - 4)" and "nkeq2 \<equiv> ((\<Sum>i::nat=0..(n-1). (1 / k i)) = 1)" shows "(nkpos \<and> nkeq1 \<and> nkeq2) \<longleftrightarrow> (n, k) \<in> putnam_2005_b2_solution" sorry end

putnam_2005_b3

abbrev putnam_2005_b3_solution : Set (Set.Ioi (0 : ℝ) → Set.Ioi (0 : ℝ)) := sorry -- {f : Set.Ioi (0 : ℝ) → Set.Ioi (0 : ℝ) | ∃ c d : ℝ, c > 0 ∧ d > 0 ∧ (d = 1 → c = 1) ∧ (∀ x : Set.Ioi (0 : ℝ), f x = c * x ^ d)} theorem putnam_2005_b3 (f : ℝ → ℝ) (fexa : Prop) (hfexa : fexa = (∃ a > 0, ∀ x > 0, deriv f (a / x) = x / f x)) : (DifferentiableOn ℝ f (Set.Ioi (0 : ℝ)) ∧ fexa) ↔ (∃ f' ∈ putnam_2005_b3_solution, ∀ x : Set.Ioi (0 : ℝ), f' x = f x) := sorry

Find all differentiable functions $f:(0,\infty) \to (0,\infty)$ for which there is a positive real number $a$ such that $f'(\frac{a}{x})=\frac{x}{f(x)}$ for all $x>0$.

Show that the functions are precisely $f(x)=cx^d$ for $c,d>0$ arbitrary except that we must take $c=1$ in case $d=1$.

['analysis']

Section putnam_2005_b3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2005_b3_solution (f: R -> R) := exists (c d: R), c > 0 /\ d > 0 /\ f = (fun x => c * Rpower x d). Theorem putnam_2005_b3: forall (f: R -> R) (x: R), x > 0 /\ f x > 0 /\ ex_derive f x -> exists (a: R), Derive f (a / x) = x / f x <-> putnam_2005_b3_solution f. Proof. Admitted. End putnam_2005_b3.

theory putnam_2005_b3 imports Complex_Main "HOL-Analysis.Derivative" begin (* uses (real \<Rightarrow> real) instead of ({0<..} \<Rightarrow> {0<..}) *) definition putnam_2005_b3_solution :: "(real \<Rightarrow> real) set" where "putnam_2005_b3_solution \<equiv> undefined" (* {f::real\<Rightarrow>real. (\<exists>c d::real. c > 0 \<and> d > 0 \<and> (d = 1 \<longrightarrow> c = 1) \<and> (\<forall>x::real\<in>{0<..}. f x = c * x powr d))} *) theorem putnam_2005_b3: fixes f :: "real \<Rightarrow> real" and fexa :: bool assumes hfexa: "fexa \<equiv> (\<exists>a::real>0. \<forall>x::real>0. deriv f (a/x) = x / f x)" shows "(f differentiable_on {0<..} \<and> fexa) \<longleftrightarrow> f \<in> putnam_2005_b3_solution" sorry end

putnam_2005_b4

theorem putnam_2005_b4 (m n : ℤ) (mnpos : m > 0 ∧ n > 0) (f : ℤ → ℤ → ℕ) (hf : ∀ m' > 0, ∀ n' > 0, f m' n' = Set.encard {x : Finset.Icc 1 n' → ℤ | ∑ i : Finset.Icc 1 n', |x i| ≤ m'}) : f m n = f n m := sorry

For positive integers $m$ and $n$, let $f(m,n)$ denote the number of $n$-tuples $(x_1,x_2,\dots,x_n)$ of integers such that $|x_1|+|x_2|+\cdots+|x_n| \leq m$. Show that $f(m,n)=f(n,m)$.

['algebra']

Section putnam_2005_b4. Require Import List Ensembles Finite_sets ZArith. Open Scope Z. Theorem putnam_2005_b4: let fix absl (l : list Z) : list Z := match l with | nil => nil | h :: t => Z.abs h :: t end in forall (m n: nat), forall (E1 E2: Ensemble (list Z)) (l1 l2: list Z), length l1 = n /\ length l2 = m /\ (E1 l1 <-> fold_left Z.add (absl l1) 0 <= Z.of_nat m) /\ (E2 l2 <-> fold_left Z.add (absl l2) 0 <= Z.of_nat n) <-> exists (a: nat), cardinal (list Z) E1 a /\ cardinal (list Z) E2 a. Proof. Admitted. End putnam_2005_b4.

theory putnam_2005_b4 imports Complex_Main begin theorem putnam_2005_b4: fixes m n::nat and f::"nat\<Rightarrow>nat\<Rightarrow>nat" assumes mnpos : "m > 0 \<and> n > 0" and "\<forall>m' > 0. \<forall>n' > 0. f m' n' = card {x::int list. size x = n' \<and> (\<Sum>i=0..<n'. abs(x!i)) \<le> m'}" shows "f m n = f n m" sorry end

putnam_2005_b6

theorem putnam_2005_b6 (n : ℕ) (v : Equiv.Perm (Fin n) → ℕ) (npos : n ≥ 1) (hv : ∀ p : Equiv.Perm (Fin n), v p = Set.encard {i : Fin n | p i = i}) : (∑ p : Equiv.Perm (Fin n), (Equiv.Perm.signAux p : ℤ) / (v p + 1 : ℝ)) = (-1) ^ (n + 1) * (n / (n + 1 : ℝ)) := sorry

Let $S_n$ denote the set of all permutations of the numbers $1,2,\dots,n$. For $\pi \in S_n$, let $\sigma(\pi)=1$ if $\pi$ is an even permutation and $\sigma(\pi)=-1$ if $\pi$ is an odd permutation. Also, let $\nu(\pi)$ denote the number of fixed points of $\pi$. Show that $\sum_{\pi \in S_n} \frac{\sigma(\pi)}{\nu(\pi)+1}=(-1)^{n+1}\frac{n}{n+1}$.

['linear_algebra', 'algebra']

theory putnam_2005_b6 imports Complex_Main "HOL-Combinatorics.Permutations" begin theorem putnam_2005_b6: fixes n::nat and S::"(nat\<Rightarrow>nat) set" and \<sigma>::"(nat\<Rightarrow>nat) \<Rightarrow> int" and v::"(nat\<Rightarrow>nat) \<Rightarrow> int" defines "S \<equiv> {p. p permutes {1..n}}" and "\<sigma> \<equiv> \<lambda>p. if (evenperm p) then 1 else -1" and "v \<equiv> \<lambda>p. card {x::nat. x > 0 \<and> x \<le> n \<and> p x = x}" assumes npos : "n > 0" shows "(\<Sum>p \<in> S. rat_of_int (\<sigma> p) / (rat_of_int (v p) + 1)) = (-1)^(n+1) * (rat_of_nat n) / (rat_of_nat n+1)" sorry end

putnam_1966_a1

theorem putnam_1966_a1 (f : ℤ → ℤ := fun n : ℤ => ∑ m in Finset.Icc 1 n, (if Even m then m / 2 else (m - 1)/2)) : ∀ x y : ℤ, x > 0 ∧ y > 0 ∧ x > y → x * y = f (x + y) - f (x - y) := sorry

[]

theory putnam_1966_a1 imports Complex_Main begin theorem putnam_1966_a1: fixes f :: "int \<Rightarrow> int" defines "f \<equiv> \<lambda> n :: int. \<Sum> m = 1..n. (if even m then m div 2 else (m - 1) div 2)" shows "\<forall> x y :: int. (x > 0 \<and> y > 0 \<and> x > y) \<longrightarrow> x * y = f (x + y) - f (x - y)" sorry end

putnam_1966_a3

theorem putnam_1966_a3 (x : ℕ → ℝ) (hx1 : 0 < x 1 ∧ x 1 < 1) (hxi : ∀ n ≥ 1, x (n + 1) = (x n) * (1 - (x n))) : Tendsto (fun n : ℕ => n * (x n)) ⊤ (𝓝 1) := sorry

[]

theory putnam_1966_a3 imports Complex_Main begin theorem putnam_1966_a3: fixes x :: "nat \<Rightarrow> real" assumes hx1: "0 < x 1 \<and> x 1 < 1" and hxi: "\<forall> n \<ge> 1. x (n + 1) = x n * (1 - x n)" shows "(\<lambda> n. n * x n) \<longlonglongrightarrow> 1" sorry end

putnam_1966_a4

theorem putnam_1966_a4 (a : ℕ → ℤ) (ha1 : a 1 = 2) (hai : ∀ n ≥ 1, a (n + 1) = (if ∃ m : ℤ, m^2 = a n + 1 = True then a n + 2 else a n + 1)) : ∀ n ≥ 1, a n = n + round (Real.sqrt n) := sorry

[]

theory putnam_1966_a4 imports Complex_Main begin theorem putnam_1966_a4: fixes a :: "nat \<Rightarrow> int" assumes ha1: "a 1 = 2" and hai: "\<forall> n \<ge> 1. a (n + 1) = (if (\<exists> m :: int. m ^ 2 = a n + 1) then a n + 2 else a n + 1)" shows "\<forall> n \<ge> 1. a n = n + round (sqrt n)" sorry end

putnam_1966_a5

theorem putnam_1966_a5 (C : Set (ℝ → ℝ) := {f : ℝ → ℝ | Continuous f}) (T : (ℝ → ℝ) → (ℝ → ℝ)) (linearT : ∀ a b : ℝ, ∀ f ∈ C, ∀ g ∈ C, T ((fun x => a)*f + (fun x => b)*g) = (fun x => a)*(T f) + (fun x => b)*(T g)) (localT : ∀ r s : ℝ, r ≤ s → ∀ f ∈ C, ∀ g ∈ C, (∀ x ∈ Set.Icc r s, f x = g x) → (∀ x ∈ Set.Icc r s, T f x = T g x)) : ∃ f ∈ C, ∀ g ∈ C, T g = f * g := sorry

[]

theory putnam_1966_a5 imports Complex_Main "HOL-Library.Extended_Real" begin theorem putnam_1966_a5: fixes C :: "(real \<Rightarrow> real) set" and T :: "(real \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real)" defines "C \<equiv> {f :: real \<Rightarrow> real. continuous_on UNIV f}" assumes imageTC: "\<forall> f \<in> C. T f \<in> C" and linearT: "\<forall> a b :: real. \<forall> f \<in> C. \<forall> g \<in> C. T (\<lambda> x. a * f x + b * g x) = (\<lambda> x. a * T f x + b * T g x)" and localT: "\<forall> (r :: ereal) (s :: ereal). r < s \<longrightarrow> (\<forall> f \<in> C. \<forall> g \<in> C. (\<forall> x :: real. r < x \<and> x < s \<longrightarrow> f x = g x) \<longrightarrow> (\<forall> x :: real. r < x \<and> x < s \<longrightarrow> T f x = T g x))" shows "\<exists> f \<in> C. \<forall> g \<in> C. T g = (\<lambda> x. f x * g x)" sorry end

putnam_1966_a6

theorem putnam_1966_a6 (a : ℕ → (ℕ → ℝ)) (ha : ∀ n ≥ 1, a n n = n ∧ ∀ m ≥ 1, m < n → a n m = m * Real.sqrt (1 + a n (m + 1))) : Tendsto (fun n => a n 1) ⊤ (𝓝 3) := sorry

[]

theory putnam_1966_a6 imports Complex_Main begin theorem putnam_1966_a6: fixes a :: "nat \<Rightarrow> nat \<Rightarrow> real" assumes ha: "\<forall> n \<ge> 1. a n n = n \<and> (\<forall> m \<ge> 1. m < n \<longrightarrow> a n m = m * sqrt (1 + a n (m + 1)))" shows "(\<lambda> n. a n 1) \<longlonglongrightarrow> 3" sorry end

putnam_1966_b2

theorem putnam_1966_b2 (S : ℤ → Set ℤ := fun n : ℤ => {n, n + 1, n + 2, n + 3, n + 4, n + 5, n + 6, n + 7, n + 8, n + 9}) : ∀ n : ℤ, ∃ k ∈ S n, ∀ m ∈ S n, k ≠ m → IsCoprime m k := sorry

[]

theory putnam_1966_b2 imports Complex_Main begin theorem putnam_1966_b2: fixes S :: "int \<Rightarrow> int set" defines "S \<equiv> \<lambda> n :: int. {n, n + 1, n + 2, n + 3, n + 4, n + 5, n + 6, n + 7, n + 8, n + 9}" shows "\<forall> n :: int. \<exists> k \<in> S n. \<forall> m \<in> S n. k \<noteq> m \<longrightarrow> coprime m k" sorry end

putnam_1966_b3

theorem putnam_1966_b3 (p : ℕ → ℝ) (hpos : ∀ n : ℕ, p n > 0) (hconv : ∃ r : ℝ, Tendsto (fun m : ℕ => ∑ n in Finset.Icc 1 m, 1/(p n)) ⊤ (𝓝 r)) : ∃ r : ℝ, Tendsto (fun m : ℕ => ∑ n in Finset.Icc 1 m, (p n) * n^2/(∑ i in Finset.Icc 1 n, p i)^2) ⊤ (𝓝 r) := sorry

[]

theory putnam_1966_b3 imports Complex_Main begin theorem putnam_1966_b3: fixes p :: "nat \<Rightarrow> real" assumes hpos: "\<forall> n :: nat. p n > 0" and hconv: "convergent (\<lambda> m. \<Sum> n = 1..m. 1 / p n)" shows "convergent (\<lambda> m. \<Sum> n = 1..m. (p n) * n ^ 2 / (\<Sum> i = 1..n. p i) ^ 2)" sorry end

putnam_1966_b4

theorem putnam_1966_b4 (m n : ℕ) (a : Finset ℕ) (hS : (∀ i ∈ a, i > 0) ∧ a.card = m * n + 1) : ∃ b ⊆ a, (b.card = m + 1 ∧ ∀ j ∈ b, ∀ i ∈ b, i ≠ j → ¬(j ∣ i)) ∨ (b.card = n + 1 ∧ ∀ j ∈ b, j ≠ sSup b → j ∣ sInf {i ∈ b | i > j}) := sorry

[]

theory putnam_1966_b4 imports Complex_Main begin theorem putnam_1966_b4: fixes m n :: nat and a :: "nat set" assumes ha: "(\<forall> i \<in> a. i > 0) \<and> card a = m * n + 1" shows "\<exists> b \<subseteq> a. (card b = m + 1 \<and> (\<forall> j \<in> b. \<forall> i \<in> b. i \<noteq> j \<longrightarrow> \<not>(j dvd i)) \<or> (card b = n + 1 \<and> (\<forall> j \<in> b. \<forall> i \<in> b. j < i \<longrightarrow> j dvd i)))" sorry end

putnam_1966_b6

theorem putnam_1966_b6 (y : ℝ → ℝ) (hy : Differentiable ℝ y ∧ Differentiable ℝ (deriv y)) (diffeq : deriv (deriv y) + Real.exp * y = 0) : ∃ r s N : ℝ, ∀ x > N, r ≤ y x ∧ y x ≤ s := sorry

[]

theory putnam_1966_b6 imports Complex_Main "HOL-Analysis.Derivative" begin theorem putnam_1966_b6: fixes y :: "real \<Rightarrow> real" assumes hy: "y differentiable_on UNIV \<and> (deriv y) differentiable_on UNIV" and diffeq: "\<forall> x :: real. (deriv (deriv y)) x + (exp x) * (y x) = 0" shows "\<exists> r s N :: real. \<forall> x > N. r \<le> y x \<and> y x \<le> s" sorry end

putnam_2001_a1

theorem putnam_2001_a1 (S : Type*) [Mul S] (hS : ∀ a b : S, (a * b) * a = b) : ∀ a b : S, a * (b * a) = b := sorry

Consider a set $S$ and a binary operation $*$, i.e., for each $a,b\in S$, $a*b\in S$. Assume $(a*b)*a=b$ for all $a,b\in S$. Prove that $a*(b*a)=b$ for all $a,b\in S$.

['abstract_algebra']

theory putnam_2001_a1 imports Complex_Main begin theorem putnam_2001_a1: fixes op :: "'a => 'a => 'a" assumes hop : "\<forall>a b :: 'a. op (op a b) a = b" shows "\<forall>a b :: 'a. op a (op b a) = b" sorry end

putnam_2001_a3

abbrev putnam_2001_a3_solution : Set ℤ := sorry -- {m : ℤ | ∃ k : ℤ, k^2 = m ∨ 2*k^2 = m} theorem putnam_2001_a3 (P : ℤ → Polynomial ℤ := fun m : ℤ => (Polynomial.X)^4 - (Polynomial.C (2*m + 4))*(Polynomial.X)^2 + Polynomial.C ((m - 2)^2)) : {m : ℤ | ∃ a : Polynomial ℤ, ∃ b : Polynomial ℤ, P m = a * b ∧ (∃ n ∈ Ici 1, a.coeff n ≠ 0) ∧ (∃ n ∈ Ici 1, b.coeff n ≠ 0)} = putnam_2001_a3_solution := sorry

For each integer $m$, consider the polynomial \[P_m(x)=x^4-(2m+4)x^2+(m-2)^2.\] For what values of $m$ is $P_m(x)$ the product of two non-constant polynomials with integer coefficients?

$P_m(x)$ factors into two nonconstant polynomials over the integers if and only if $m$ is either a square or twice a square.

['algebra']

Section putnam_2001_a3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2001_a3_solution (m: Z) := exists (n: Z), m = Z.mul n n \/ m = Z.mul 2 (Z.mul n n). Theorem putnam_2001_a3: let P (m: Z) (x: R) := x ^ 4 - (2 * IZR m + 4) * x ^ 2 + (IZR m - 2) ^ 2 in let p (a: nat -> Z) (x: R) (n: nat) := sum_n (fun i => IZR (a i) * x ^ i) n in forall (m: Z), exists (a1 a2: nat -> Z) (n1 n2: nat), forall (x: R), P m x = p a1 x n1 * p a2 x n2 <-> putnam_2001_a3_solution m. Proof. Admitted. End putnam_2001_a3.

theory putnam_2001_a3 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_2001_a3_solution :: "int set" where "putnam_2001_a3_solution \<equiv> undefined" (* {m :: int. \<exists> k :: int. k ^ 2 = m \<or> 2 * k ^ 2 = m} *) theorem putnam_2001_a3: fixes P :: "int \<Rightarrow> int poly" defines "P \<equiv> \<lambda> m :: int. monom 1 4 - monom (2 * m + 4) 2 + monom ((m - 2) ^ 2) 0" shows "{m :: int. \<exists> a b :: int poly. P m = a * b \<and> (\<exists> n \<in> {1..}. coeff a n \<noteq> 0) \<and> (\<exists> n \<in> {1..}. coeff b n \<noteq> 0)} = putnam_2001_a3_solution" sorry end

putnam_2001_a5

theorem putnam_2001_a5 : ∃! (a : ℤ) (n : ℕ), a > 0 ∧ n > 0 ∧ a^(n+1) - (a+1)^n = 2001 := sorry

Prove that there are unique positive integers $a$, $n$ such that $a^{n+1}-(a+1)^n=2001$.

['number_theory']

Section putnam_2001_a5. Require Import Nat. Theorem putnam_2001_a5 : exists! (a n: nat), a > 0 /\ n > 0 /\ a ^ (n + 1) - (a + 1) ^ n = 2001. Proof. Admitted. End putnam_2001_a5.

theory putnam_2001_a5 imports Complex_Main begin theorem putnam_2001_a5: shows "\<exists>! (a :: int). \<exists>! (n :: nat). a > 0 \<and> \<and> n > 0 \<and> a^(n+1) - (a+1)^n = 2001" sorry end

putnam_2001_b1

theorem putnam_2001_b1 (n : ℕ) (nums : Fin n → Fin n → ℤ) (colors : Fin n → Fin n → Fin 2) (npos : n > 0) (neven : Even n) (hnums : ∀ k l : Fin n, nums k l = k * n + l + 1) (hcolorsrows : ∀ k : Fin n, (∑ l : Fin n, (if (colors k l = 0) then 1 else 0)) = n / 2) (hcolorscols : ∀ l : Fin n, (∑ k : Fin n, (if (colors k l = 0) then 1 else 0)) = n / 2) : (∑ k : Fin n, ∑ l : Fin n, (if (colors k l = 0) then (nums k l) else 0)) = (∑ k : Fin n, ∑ l : Fin n, (if (colors k l = 1) then (nums k l) else 0)) := sorry

Let $n$ be an even positive integer. Write the numbers $1,2,\ldots,n^2$ in the squares of an $n \times n$ grid so that the $k$-th row, from left to right, is $(k-1)n+1,(k-1)n+2,\ldots,(k-1)n+n$. Color the squares of the grid so that half of the squares in each row and in each column are red and the other half are black (a checkerboard coloring is one possibility). Prove that for each coloring, the sum of the numbers on the red squares is equal to the sum of the numbers on the black squares.

['algebra']

theory putnam_2001_b1 imports Complex_Main begin theorem putnam_2001_b1: fixes n :: nat and nums :: "nat \<Rightarrow> nat \<Rightarrow> nat" and colors :: "nat \<Rightarrow> nat \<Rightarrow> nat" assumes npos: "n > 0" and neven: "even n" and hnums: "\<forall> k \<in> {0 ..< n}. \<forall> l \<in> {0 ..< n}. nums k l = k * n + l + 1" and hcolorsrng: "\<forall> k \<in> {0 ..< n}. \<forall> l \<in> {0 ..< n}. colors k l \<in> {0, 1}" and hcolorsrows: "\<forall> k \<in> {0 ..< n}. (\<Sum> l = 0 ..< n. colors k l) = n div 2" and hcolorscols: "\<forall> l \<in> {0 ..< n}. (\<Sum> k = 0 ..< n. colors k l) = n div 2" shows "(\<Sum> k = 0 ..< n. \<Sum> l = 0 ..< n. if colors k l = 0 then nums k l else 0) = (\<Sum> k = 0 ..< n. \<Sum> l = 0 ..< n. if colors k l = 1 then nums k l else 0)" sorry end

putnam_2001_b2

abbrev putnam_2001_b2_solution : Set (ℝ × ℝ) := sorry -- {((3 ^ ((1 : ℝ) / 5) + 1) / 2, (3 ^ ((1 : ℝ) / 5) - 1) / 2)} theorem putnam_2001_b2 (x y : ℝ) (eq1 : Prop) (eq2 : Prop) (heq1 : eq1 = (1 / x + 1 / (2 * y) = (x ^ 2 + 3 * y ^ 2) * (3 * x ^ 2 + y ^ 2))) (heq2 : eq2 = (1 / x - 1 / (2 * y) = 2 * (y ^ 4 - x ^ 4))) : (eq1 ∧ eq2) ↔ (x, y) ∈ putnam_2001_b2_solution := sorry

Find all pairs of real numbers $(x,y)$ satisfying the system of equations \begin{align*} \frac{1}{x}+\frac{1}{2y}&=(x^2+3y^2)(3x^2+y^2) \\ \frac{1}{x}-\frac{1}{2y}&=2(y^4-x^4). \end{align*}

Show that $x=(3^{1/5}+1)/2$ and $y=(3^{1/5}-1)/2$ is the unique solution satisfying the given equations.

['algebra']

Section putnam_2001_b2. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2001_b2_solution (x y: R) := x = (3 ^ (1 / 5) + 1) / 2 /\ y = (3 ^ (1 / 5) - 1) / 2. Theorem putnam_2001_b2: forall (x y: R), 1 / x + 1 / (2 * y) = (x ^ 2 + 3 * y ^ 2) * (3 * x ^ 2 + y ^ 2) /\ 1 / x - 1 / (2 * y) = 2 * (y ^ 4 - x ^ 4) <-> putnam_2001_b2_solution x y. Proof. Admitted. End putnam_2001_b2.

theory putnam_2001_b2 imports Complex_Main begin definition putnam_2001_b2_solution :: "(real \<times> real) set" where "putnam_2001_b2_solution \<equiv> undefined" (* {((3 powr (1 / 5) + 1) / 2, (3 powr (1 / 5) - 1) / 2)} *) theorem putnam_2001_b2: fixes x y :: real and eq1 eq2 :: bool defines "eq1 \<equiv> 1 / x + 1 / (2 * y) = (x ^ 2 + 3 * y ^ 2) * (3 * x ^ 2 + y ^ 2)" and "eq2 \<equiv> 1 / x - 1 / (2 * y) = 2 * (y ^ 4 - x ^ 4)" shows "(eq1 \<and> eq2) \<longleftrightarrow> (x, y) \<in> putnam_2001_b2_solution" sorry end

putnam_2001_b3

abbrev putnam_2001_b3_solution : ℝ := sorry -- 3 theorem putnam_2001_b3 : ∑' n : Set.Ici 1, ((2 : ℝ) ^ (round (Real.sqrt n)) + (2 : ℝ) ^ (-round (Real.sqrt n))) / 2 ^ (n : ℝ) = putnam_2001_b3_solution := sorry

For any positive integer $n$, let $\langle n \rangle$ denote the closest integer to $\sqrt{n}$. Evaluate $\sum_{n=1}^\infty \frac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}$.

Show that the sum is $3$.

['analysis']

Section putnam_2001_b3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2001_b3_solution := 3. Theorem putnam_2001_b3: forall (n: nat), ge n 0 -> let closest (n: nat) := let n := INR n in Rmin ((n - sqrt n) - IZR (floor (n - sqrt n))) (IZR (floor (n + 1 - sqrt n)) - (n - sqrt n)) in Lim_seq (fun n => sum_n (fun n => (Rpower 2 (closest n) + Rpower 2 (- closest n)) / (2 ^ n)) n) = putnam_2001_b3_solution. Proof. Admitted. End putnam_2001_b3.

theory putnam_2001_b3 imports Complex_Main begin definition putnam_2001_b3_solution :: real where "putnam_2001_b3_solution \<equiv> undefined" (* 3 *) theorem putnam_2001_b3: shows "(\<Sum> n :: nat. (2 powr (round (sqrt (n + 1))) + 2 powr (- round (sqrt (n + 1)))) / 2 powr (n + 1)) = putnam_2001_b3_solution" sorry end

putnam_2001_b4

abbrev putnam_2001_b4_solution : Prop := sorry -- True theorem putnam_2001_b4 (S : Set ℚ) (hS : S = univ \ {-1, 0, 1}) (f : S → S) (hf : ∀ x : S, f x = x - 1 / (x : ℚ)) : ⋂ n ∈ Ici 1, f^[n] '' univ = ∅ ↔ putnam_2001_b4_solution := sorry

Let $S$ denote the set of rational numbers different from $\{-1,0,1\}$. Define $f:S\rightarrow S$ by $f(x)=x-1/x$. Prove or disprove that \[\bigcap_{n=1}^\infty f^{(n)}(S) = \emptyset,\] where $f^{(n)}$ denotes $f$ composed with itself $n$ times.

['algebra']

theory putnam_2001_b4 imports Complex_Main begin definition putnam_2001_b4_solution::bool where "putnam_2001_b4_solution \<equiv> undefined" (* True *) theorem putnam_2001_b4: fixes f :: "rat \<Rightarrow> rat" and S :: "rat set" defines "f \<equiv> \<lambda>x. x - 1/x" and "S \<equiv> \<rat> - {-1, 0, 1}" shows "putnam_2001_b4_solution = ((\<Inter>n \<in> {1..}. image (f^^n) S) = {}) " sorry end

putnam_2001_b5

theorem putnam_2001_b5 (a b : ℝ) (g : ℝ → ℝ) (abint : 0 < a ∧ a < 1 / 2 ∧ 0 < b ∧ b < 1 / 2) (gcont : Continuous g) (hg : ∀ x : ℝ, g (g x) = a * g x + b * x) : ∃ c : ℝ, ∀ x : ℝ, g x = c * x := sorry

Let $a$ and $b$ be real numbers in the interval $(0,1/2)$, and let $g$ be a continuous real-valued function such that $g(g(x))=ag(x)+bx$ for all real $x$. Prove that $g(x)=cx$ for some constant $c$.

['analysis']

Section putnam_2001_b5. Require Import Reals Coquelicot.Coquelicot. Theorem putnam_2001_b5: forall (a b: R) (g: R -> R), 0 < a < 1/2 /\ 0 < b < 1/2 /\ continuity g /\ forall (x: R), g (g x) = a * g x + b * x -> exists (c: R), g x = c * x. Proof. Admitted. End putnam_2001_b5.

theory putnam_2001_b5 imports Complex_Main begin theorem putnam_2001_b5: fixes a b :: real and g :: "real \<Rightarrow> real" assumes abint: "a \<in> {0 <..< 1 / 2} \<and> b \<in> {0 <..< 1 / 2}" and gcont: "continuous_on UNIV g" and hg : "\<forall> x :: real. g (g x) = a * g x + b * x" shows "\<exists> c :: real. \<forall> x :: real. g x = c * x" sorry end

putnam_2001_b6

abbrev putnam_2001_b6_solution : Prop := sorry -- True theorem putnam_2001_b6 (aposinc : (ℤ → ℝ) → Prop) (alim : (ℤ → ℝ) → Prop) (haposinc : ∀ a : ℤ → ℝ, aposinc a = ∀ n ≥ 1, a n > 0 ∧ a n < a (n + 1)) (halim : ∀ a : ℤ → ℝ, alim a = Tendsto (fun n : ℤ => a (n + 1) / (n + 1)) atTop (𝓝 0)) : (∀ a : ℤ → ℝ, (aposinc a ∧ alim a) → {n : ℤ | n > 0 ∧ (∀ i ∈ Set.Icc 1 (n - 1), a (n - i) + a (n + i) < 2 * a n)}.Infinite) ↔ putnam_2001_b6_solution := sorry

Assume that $(a_n)_{n \geq 1}$ is an increasing sequence of positive real numbers such that $\lim a_n/n=0$. Must there exist infinitely many positive integers $n$ such that $a_{n-i}+a_{n+i}<2a_n$ for $i=1,2,\ldots,n-1$?

Show that the answer is yes, there must exist infinitely many such $n$.

['analysis']

Section putnam_2001_b6. Require Import Nat Reals Coquelicot.Coquelicot. Definition putnam_2001_b6_solution := True. Theorem putnam_2001_b6: forall (a: nat -> R), (forall (i j: nat), lt i j -> a i < a j) /\ Lim_seq (fun n => a n / INR n) = 0 -> forall (n: nat), exists (m: nat), gt m n /\ forall (i: nat), and (le 1 i) (le i (n - 1)) -> a (sub n i) + a (add n i) < 2 * a n /\ a (sub m i) + a (add m i) < 2 * a m. Proof. Admitted. End putnam_2001_b6.

theory putnam_2001_b6 imports Complex_Main begin definition putnam_2001_b6_solution :: bool where "putnam_2001_b6_solution \<equiv> undefined" (* True *) theorem putnam_2001_b6: fixes aposinc alim :: "(nat \<Rightarrow> real) \<Rightarrow> bool" defines "aposinc \<equiv> \<lambda> a. \<forall> n \<ge> 1. a n > 0 \<and> a n < a (n + 1)" and "alim \<equiv> \<lambda> a. (\<lambda> n :: nat. a (n + 1) / (n + 1)) \<longlonglongrightarrow> 0" shows "(\<forall> a :: nat \<Rightarrow> real. (aposinc a \<and> alim a) \<longrightarrow> infinite {n :: nat. n > 0 \<and> (\<forall> i \<in> {1 .. n - 1}. a (n - i) + a (n + i) < 2 * a n)}) \<longleftrightarrow> putnam_2001_b6_solution" sorry end

putnam_1968_a1

theorem putnam_1968_a1 : 22/7 - Real.pi = ∫ x in (0)..1, x^4 * (1 - x)^4 / (1 + x^2) := sorry

[]

theory putnam_1968_a1 imports Complex_Main "HOL-Analysis.Interval_Integral" begin theorem putnam_1968_a1: shows "22 / 7 - pi = interval_lebesgue_integral lebesgue 0 1 (\<lambda> x. x ^ 4 * (1 - x) ^ 4 / (1 + x ^ 2))" sorry end

putnam_1968_a2

theorem putnam_1968_a2 (a b c d e f : ℤ) (ε : ℝ) (hne : a * d ≠ b * c) (hε : ε > 0) (q1 : ℚ := |r * a + s * b - e|) (q2 : ℚ := |r * c + s * d - f|) : ∃ r s : ℚ, 0 < q1 ∧ q1 < ε ∧ 0 < q2 ∧ q2 < ε := sorry

[]

theory putnam_1968_a2 imports Complex_Main begin theorem putnam_1968_a2: fixes a b c d e f :: int and \<epsilon> :: real assumes hne: "a * d \<noteq> b * c" and h\<epsilon>: "\<epsilon> > 0" shows "\<exists> r s :: rat. 0 < real_of_rat \<bar>r * rat_of_int a + s * rat_of_int b - rat_of_int e\<bar> \<and> real_of_rat \<bar>r * rat_of_int a + s * rat_of_int b - rat_of_int e\<bar> < \<epsilon> \<and> 0 < real_of_rat \<bar>r * rat_of_int c + s * rat_of_int d - rat_of_int f\<bar> \<and> real_of_rat \<bar>r * rat_of_int c + s * rat_of_int d - rat_of_int f\<bar> < \<epsilon>" sorry end

putnam_1968_a3

theorem putnam_1968_a3 {a : Type} (S : Finset a) (ha : SDiff (Finset a)) : ∃ l : List (Finset a), (∀ i : Fin l.length, l[i] ⊆ S) ∧ l[0]! = ∅ ∧ (∀ s ⊆ S, ∃! i : Fin l.length, l[i] = s) ∧ ∀ i ∈ Finset.range (l.length - 1), (l[i]! ⊆ l[i+1]! ∧ (l[i+1]! \ l[i]!).card = 1) ∨ (l[i+1]! ⊆ l[i]! ∧ (l[i]! \ l[i+1]!).card = 1) := sorry

[]

theory putnam_1968_a3 imports Complex_Main begin theorem putnam_1968_a3: shows "\<exists> l :: ('a::finite) set list. l!0 = {} \<and> (\<forall> S :: 'a set. \<exists>! i \<in> {0 ..< length l}. l!i = S) \<and> (\<forall> i \<in> {0 ..< length l - 1}. \<exists> a :: 'a. (a \<notin> l!i \<and> l!(i+1) = insert a (l!i)) \<or> (a \<notin> l!(i+1) \<and> l!i = insert a (l!(i+1))))" sorry end

putnam_1968_a4

theorem putnam_1968_a4 (n : ℕ) (S : Fin n → (Fin 3 → ℝ)) (hS : ∀ i : Fin n, (S i 0)^2 + (S i 1)^2 + (S i 2)^2 = 1) : ∑ i : Fin n, ∑ j : Fin n, (if i < j then (Euclidean.dist (S i) (S j))^2 else (0 : ℝ)) ≤ n^2 := sorry

[]

theory putnam_1968_a4 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin (* Note: This formalization uses "nat" instead of "Fin n" as the domain of S *) theorem putnam_1968_a4: fixes n :: nat and S :: "nat \<Rightarrow> real^3" assumes hS: "\<forall> i \<in> {0..<n}. (S i $ 1)^2 + (S i $ 2)^2 + (S i $ 3)^2 = 1" shows "(\<Sum> i = 0..<n. \<Sum> j = 0..<n. if i < j then (dist (S i) (S j)) ^ 2 else 0) \<le> n ^ 2" sorry end

putnam_1968_a5

abbrev putnam_1968_a5_solution : ℝ := sorry -- 8 theorem putnam_1968_a5 (V : Set ℝ[X] := {P : ℝ[X] | P.degree = 2 ∧ ∀ x ∈ Set.Icc 0 1, |P.eval x| ≤ 1}) : sSup {|(derivative P).eval 0| | P ∈ V} = putnam_1968_a5_solution := sorry

[]

theory putnam_1968_a5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1968_a5_solution :: real where "putnam_1968_a5_solution \<equiv> undefined" (* 8 *) theorem putnam_1968_a5: fixes V :: "real poly set" defines "V \<equiv> {P :: real poly. degree P = 2 \<and> (\<forall> x \<in> {0..1}. \<bar>poly P x\<bar> \<le> 1)}" shows "Sup {y :: real. \<exists> P \<in> V. \<bar>poly (pderiv p) 0\<bar> = y} = putnam_1968_a5_solution" sorry end

putnam_1968_a6

abbrev putnam_1968_a6_solution : Set ℂ[X] := sorry -- {X - 1, -(X - 1), X + 1, -(X + 1), X^2 + X - 1, -(X^2 + X - 1), X^2 - X - 1, -(X^2 - X - 1), X^3 + X^2 - X - 1, -(X^3 + X^2 - X - 1), X^3 - X^2 - X + 1, -(X^3 - X^2 - X + 1)} theorem putnam_1968_a6 : {P : ℂ[X] | P.natDegree ≥ 1 ∧ (∀ k ∈ Set.Icc 0 P.natDegree, P.coeff k = 1 ∨ P.coeff k = -1) ∧ ∀ z : ℂ, P.eval z = 0 → ∃ r : ℝ, r = z} = putnam_1968_a6_solution := sorry

[]

theory putnam_1968_a6 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1968_a6_solution :: "complex poly set" where "putnam_1968_a6_solution \<equiv> undefined" (* {P :: complex poly. \<exists> l \<in> {[-1, 1], [1, 1], [-1, 1, 1], [-1, -1, 1], [-1, -1, 1, 1], [1, -1, -1, 1]}. P = Poly l \<or> P = -(Poly l)} *) theorem putnam_1968_a6: shows "{P :: complex poly. degree P \<ge> 1 \<and> (\<forall> k \<in> {0 .. degree P}. coeff P k = 1 \<or> coeff P k = -1) \<and> (\<forall> z :: complex. poly P z = 0 \<longrightarrow> (\<exists> r :: real. r = z))} = putnam_1968_a6_solution" sorry end

putnam_1968_b2

theorem putnam_1968_b2 [Group G] (hG : Finite G) (A : Set G) (hA : A.ncard > (Nat.card G : ℚ)/2) : ∀ g : G, ∃ x ∈ A, ∃ y ∈ A, g = x * y := sorry

[]

theory putnam_1968_b2 imports Complex_Main "HOL-Algebra.Multiplicative_Group" begin theorem putnam_1968_b2: fixes A :: "'a set" and G :: "'a monoid" assumes hG: "group G \<and> finite (carrier G)" and hAsG: "A \<subseteq> carrier G" and hA: "card A > card (carrier G) / 2" shows "\<forall> g \<in> carrier G. \<exists> x y. x \<in> A \<and> y \<in> A \<and> g = x \<otimes>\<^bsub>G\<^esub> y" sorry end

putnam_1968_b4

theorem putnam_1968_b4 (f : ℝ → ℝ) (hf : Continuous f ∧ ∃ r : ℝ, Tendsto (fun y => ∫ x in ball 0 y, f x) ⊤ (𝓝 r)) : ∃ r : ℝ, Tendsto (fun y => ∫ x in ball 0 y, f (x - 1/x)) ⊤ (𝓝 r) ∧ Tendsto (fun y => ∫ x in ball 0 y, f x) ⊤ (𝓝 r) := sorry

[]

theory putnam_1968_b4 imports Complex_Main "HOL-Analysis.Interval_Integral" begin theorem putnam_1968_b4: fixes f :: "real \<Rightarrow> real" assumes hf: "continuous_on UNIV f \<and> (\<exists> r :: real. ((\<lambda> y. interval_lebesgue_integral lebesgue (-y) y f) \<longlongrightarrow> r) at_top)" shows "\<exists> r :: real. ((\<lambda> y. interval_lebesgue_integral lebesgue (-y) y (\<lambda> x. f (x - 1 / x))) \<longlongrightarrow> r) at_top \<and> ((\<lambda> y. interval_lebesgue_integral lebesgue (-y) y f) \<longlongrightarrow> r) at_top" sorry end

putnam_1968_b5

abbrev putnam_1968_b5_solution : ℕ → ℕ := sorry -- fun p => p^2 + p theorem putnam_1968_b5 (p : ℕ) (hp : Prime p) : {M : Matrix (Fin 2) (Fin 2) (ZMod p) | M 0 0 + M 1 1 = 1 ∧ M 0 0 * M 1 1 - M 0 1 * M 1 0 = 0}.ncard = putnam_1968_b5_solution p := sorry

[]

theory putnam_1968_b5 imports Complex_Main "HOL-Computational_Algebra.Primes" "HOL-Analysis.Finite_Cartesian_Product" begin definition putnam_1968_b5_solution :: "nat \<Rightarrow> nat" where "putnam_1968_b5_solution \<equiv> undefined" (* \<lambda> p. p ^ 2 + p *) theorem putnam_1968_b5: fixes p :: nat assumes hp: "prime p" and pcard: "CARD('p::field) = p" shows "card {M :: 'p^2^2. M$1$1 + M$2$2 = 1 \<and> M$1$1 * M$2$2 - M$1$2 * M$2$1 = 0} = putnam_1968_b5_solution p" sorry end

putnam_1968_b6

theorem putnam_1968_b6 : ¬∃ K : ℕ → Set ℚ, (∀ n : ℕ, IsCompact (K n)) ∧ (∀ S : Set ℚ, IsCompact S → ∃ n : ℕ, S ⊆ K n) := sorry

[]

theory putnam_1968_b6 imports Complex_Main begin theorem putnam_1968_b6: fixes compactQ :: "real set \<Rightarrow> bool" defines "compactQ \<equiv> \<lambda> S. compact S \<and> (\<forall> r \<in> S. \<exists> q :: rat. q = r)" shows "\<not>(\<exists> K :: nat \<Rightarrow> real set. (\<forall> n :: nat. compactQ (K n)) \<and> (\<forall> S :: real set. compactQ S \<longrightarrow> (\<exists> n :: nat. S \<subseteq> K n)))" sorry end

putnam_1967_a1

theorem putnam_1967_a1 (n : ℕ) (a : Set.Icc 1 n → ℝ) (f : ℝ → ℝ := (fun x : ℝ => ∑ i : Set.Icc 1 n, a i * Real.sin (i * x))) (npos : n > 0) (flesin : ∀ x : ℝ, abs (f x) ≤ abs (Real.sin x)) : abs (∑ i : Set.Icc 1 n, i * a i) ≤ 1 := sorry

Let $f(x)=a_1\sin x+a_2\sin 2x+\dots+a_n\sin nx$, where $a_1,a_2,\dots,a_n$ are real numbers and where $n$ is a positive integer. Given that $|f(x)| \leq |\sin x|$ for all real $x$, prove that $|a_1|+|2a_2|+\dots+|na_n| \leq 1$.

['analysis']

theory putnam_1967_a1 imports Complex_Main begin (* uses (nat \<Rightarrow> real) instead of ({1..n} \<Rightarrow> real) *) theorem putnam_1967_a1: fixes n :: nat and a :: "nat \<Rightarrow> real" and f :: "real \<Rightarrow> real" defines "f \<equiv> (\<lambda>x::real. (\<Sum>i::nat=1..n. a i * sin (i*x)))" assumes npos: "n > 0" and flesin: "\<forall>x::real. \<bar>f x\<bar> \<le> \<bar>sin x\<bar>" shows "\<bar>\<Sum>i::nat=1..n. i * a i\<bar> \<le> 1" sorry end

putnam_1967_a2

theorem putnam_1967_a2 (S : ℕ → ℤ) (hS0 : S 0 = 1) (hSn : ∀ n ≥ 1, S n = {A : Matrix (Fin n) (Fin n) ℕ | (∀ i j : Fin n, A i j = A j i) ∧ (∀ j : Fin n, (∑ i : Fin n, A i j) = 1)}.ncard) : (∀ n ≥ 1, S (n + 1) = S n + n * S (n - 1)) ∧ (∀ x : ℝ, (∑' n : ℕ, S n * (x ^ n / (n)!)) = Real.exp (x + x ^ 2 / 2)) := sorry

Define $S_0$ to be $1$. For $n \geq 1$, let $S_n$ be the number of $n \times n$ matrices whose elements are nonnegative integers with the property that $a_{ij}=a_{ji}$, ($i,j=1,2,\dots,n$) and where $\sum_{i=1}^n a_{ij}=1$, ($j=1,2,\dots,n$). Prove \begin{enumerate} \item[(a)] $S_{n+1}=S_n+nS_{n-1}$ \item[(b)] $\sum_{n=0}^\infty S_n\frac{x^n}{n!} = \exp(x+x^2/2)$, where $\exp(x)=e^x$. \end{enumerate}

['linear_algebra', 'analysis']

theory putnam_1967_a2 imports Complex_Main begin (* uses (nat \<Rightarrow> nat \<Rightarrow> nat) instead of (Fin n \<Rightarrow> Fin n \<Rightarrow> nat) *) theorem putnam_1967_a2: fixes S :: "nat \<Rightarrow> int" assumes hS0: "S 0 = 1" and hSn: "\<forall>n::nat\<ge>1. S n = card {A::nat\<Rightarrow>nat\<Rightarrow>nat. (\<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. A i j = A j i) \<and> (\<forall>j::nat\<in>{0..(n-1)}. (\<Sum>i::nat=0..(n-1). A i j) = 1) \<and> (\<forall>i j::nat. ((i \<ge> n \<or> j \<ge> n) \<longrightarrow> A i j = 0))}" shows "(\<forall>n::nat\<ge>1. S (n+1) = S n + n * S (n-1)) \<and> (\<forall>x::real. (\<Sum>n::nat. S n * (x^n / fact n)) = exp (x + x^2/2))" sorry end

putnam_1967_a3

abbrev putnam_1967_a3_solution : ℕ := sorry -- 5 theorem putnam_1967_a3 (pform : Polynomial ℝ → Prop := (fun p : Polynomial ℝ => p.degree = 2 ∧ ∀ i ∈ Finset.range 3, p.coeff i = round (p.coeff i))) (pzeros : Polynomial ℝ → Prop := (fun p : Polynomial ℝ => ∃ z1 z2 : Set.Ioo (0 : ℝ) 1, z1 ≠ z2 ∧ p.eval z1.1 = 0 ∧ p.eval z2.1 = 0)) (pall : Polynomial ℝ → Prop := (fun p : Polynomial ℝ => pform p ∧ pzeros p ∧ p.coeff 2 > 0)) : (∃ p : Polynomial ℝ, pall p ∧ p.coeff 2 = putnam_1967_a3_solution) ∧ (∀ p : Polynomial ℝ, pall p → p.coeff 2 ≥ putnam_1967_a3_solution) := sorry

Consider polynomial forms $ax^2-bx+c$ with integer coefficients which have two distinct zeros in the open interval $0<x<1$. Exhibit with a proof the least positive integer value of $a$ for which such a polynomial exists.

Show that the minimum possible value for $a$ is $5$.

['algebra']

theory putnam_1967_a3 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_1967_a3_solution :: nat where "putnam_1967_a3_solution \<equiv> undefined" (* 5 *) theorem putnam_1967_a3: fixes pform :: "(real poly) \<Rightarrow> bool" and pzeros :: "(real poly) \<Rightarrow> bool" and pall :: "(real poly) \<Rightarrow> bool" defines "pform \<equiv> (\<lambda>p::real poly. degree p = 2 \<and> (\<forall>i::nat\<in>{0..2}. coeff p i \<in> \<int>))" and "pzeros \<equiv> (\<lambda>p::real poly. (\<exists>z1::real\<in>{0<..<1}. \<exists>z2::real\<in>{0<..<1}. z1 \<noteq> z2 \<and> poly p z1 = 0 \<and> poly p z2 = 0))" and "pall \<equiv> (\<lambda>p::real poly. pform p \<and> pzeros p \<and> coeff p 2 > 0)" shows "(LEAST coeff2::nat. (\<exists>p::real poly. pall p \<and> coeff2 = coeff p 2)) = putnam_1967_a3_solution" sorry end

putnam_1967_a4

theorem putnam_1967_a4 (lambda : ℝ) (hlambda : lambda > 1 / 2) : ¬∃ u : ℝ → ℝ, ∀ x ∈ Set.Icc 0 1, u x = 1 + lambda * (∫ y in Set.Ioo x 1, u y * u (y - x)) := sorry

Show that if $\lambda > \frac{1}{2}$ there does not exist a real-valued function $u$ such that for all $x$ in the closed interval $0 \leq x \leq 1$, $u(x)=1+\lambda\int_x^1 u(y)u(y-x)\,dy$.

['analysis']

theory putnam_1967_a4 imports Complex_Main "HOL-Analysis.Interval_Integral" begin theorem putnam_1967_a4: fixes lambda :: real assumes hlambda: "lambda > 1/2" shows "\<not>(\<exists>u::real\<Rightarrow>real. \<forall>x::real\<in>{0..1}. u x = 1 + lambda * (interval_lebesgue_integral lebesgue x 1 (\<lambda>y::real. u y * u (y - x))))" sorry end

putnam_1967_b2

theorem putnam_1967_b2 (p r : ℝ) (A B C α β γ : ℝ) (prbound : 0 ≤ p ∧ p ≤ 1 ∧ 0 ≤ r ∧ r ≤ 1) (id1 : ∀ x y : ℝ, (p * x + (1 - p) * y) ^ 2 = A * x ^ 2 + B * x * y + C * y ^ 2) (id2 : ∀ x y : ℝ, (p * x + (1 - p) * y) * (r * x + (1 - r) * y) = α * x ^ 2 + β * x * y + γ * y ^ 2) : max (max A B) C ≥ 4 / 9 ∧ max (max α β) γ ≥ 4 / 9 := sorry

Let $0 \leq p \leq 1$ and $0 \leq r \leq 1$ and consider the identities \begin{enumerate} \item[(a)] $(px+(1-p)y)^2=Ax^2+Bxy+Cy^2$, \item[(b)] $(px+(1-p)y)(rx+(1-r)y)=\alpha x^2+\beta xy+\gamma y^2$. \end{enumerate} Show that (with respect to $p$ and $r$) \begin{enumerate} \item[(a)] $\max\{A,B,C\} \geq 4/9$, \item[(b)] $\max\{\alpha,\beta,\gamma\} \geq 4/9$. \end{enumerate}

['algebra']

theory putnam_1967_b2 imports Complex_Main begin theorem putnam_1967_b2: fixes p r :: real and A B C \<alpha> \<beta> \<gamma> :: real assumes prbound: "0 \<le> p \<and> p \<le> 1 \<and> 0 \<le> r \<and> r \<le> 1" and id1: "\<forall>x y::real. (p*x + (1-p)*y)^2 = A*x^2 + B*x*y + C*y^2" and id2: "\<forall>x y::real. (p*x + (1-p)*y) * (r*x + (1-r)*y) = \<alpha>*x^2 + \<beta>*x*y + \<gamma>*y^2" shows "max (max A B) C \<ge> 4/9 \<and> max (max \<alpha> \<beta>) \<gamma> \<ge> 4/9" sorry end

putnam_1967_b3

theorem putnam_1967_b3 (f g : ℝ → ℝ) (fgcont : Continuous f ∧ Continuous g) (fgperiod : Function.Periodic f 1 ∧ Function.Periodic g 1) : Tendsto (fun n : ℤ => ∫ x in Set.Ioo 0 1, f x * g (n * x)) atTop (𝓝 ((∫ x in Set.Ioo 0 1, f x) * (∫ x in Set.Ioo 0 1, g x))) := sorry

If $f$ and $g$ are continuous and periodic functions with period $1$ on the real line, then $\lim_{n \to \infty} \int_0^1 f(x)g(nx)\,dx=(\int_0^1 f(x)\,dx)(\int_0^1 g(x)\,dx)$.

['analysis']

theory putnam_1967_b3 imports Complex_Main "HOL-Library.Periodic_Fun" "HOL-Analysis.Interval_Integral" begin theorem putnam_1967_b3: fixes f g :: "real \<Rightarrow> real" assumes fgcont: "continuous_on UNIV f \<and> continuous_on UNIV g" and fgperiod: "periodic_fun_simple' f \<and> periodic_fun_simple' g" shows "filterlim (\<lambda>n::int. interval_lebesgue_integral lebesgue 0 1 (\<lambda>x::real. f x * g (n*x))) (nhds ((interval_lebesgue_integral lebesgue 0 1 f) * (interval_lebesgue_integral lebesgue 0 1 g))) at_top" sorry end

putnam_2011_a2

abbrev putnam_2011_a2_solution : ℝ := sorry -- 2/Real.pi theorem putnam_2011_a2 (a b : ℕ → ℝ) (habn : ∀ n : ℕ, a n > 0 ∧ b n > 0) (hab1 : a 0 = 1 ∧ b 0 = 1) (hb : ∀ n ≥ 1, b n = b (n-1) * a n - 2) (hbnd : ∃ B : ℝ, ∀ n : ℕ, |b n| ≤ B) : Tendsto (fun n => ∑ i : Fin n, 1/(∏ j : Fin (i + 1), (a j))) atTop (𝓝 putnam_2011_a2_solution) := sorry

Let $a_1,a_2,\dots$ and $b_1,b_2,\dots$ be sequences of positive real numbers such that $a_1 = b_1 = 1$ and $b_n = b_{n-1} a_n - 2$ for$n=2,3,\dots$. Assume that the sequence $(b_j)$ is bounded. Prove tha \[ S = \sum_{n=1}^\infty \frac{1}{a_1...a_n} \] converges, and evaluate $S$.

Show that the solution is $S = 3/2$.

['analysis']

Section putnam_2011_a2. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2011_a2_solution := 3 / 2. Theorem putnam_2011_a2: let fix prod_n (m: nat -> R) (n : nat) : R := match n with | O => m 0%nat | S n' => m n' * prod_n m n' end in forall (a: nat -> R), let fix b (n: nat) := match n with | O => 1 | S n' => b n' * a n - 2 end in exists (M: nat), (forall (n: nat), a n > 0 /\ b n > 0 /\ -1 * INR M <= b n <= INR M) -> Series (fun n => prod_n a n) = putnam_2011_a2_solution. Proof. Admitted. End putnam_2011_a2.

theory putnam_2011_a2 imports Complex_Main begin definition putnam_2011_a2_solution :: real where "putnam_2011_a2_solution \<equiv> undefined" (* 2/pi *) theorem putnam_2011_a2: fixes a b :: "nat \<Rightarrow> real" assumes habn: "\<forall>n::nat. a n > 0 \<and> b n > 0" and hab1: "a 0 = 1 \<and> b 0 = 1" and hb: "\<forall>n::nat\<ge>1. b n = b (n-1) * a n - 2" and hbnd: "\<exists>B::real. \<forall>n::nat. \<bar>b n\<bar> \<le> B" shows "filterlim (\<lambda>n::nat. \<Sum>i::nat=0..(n-1). 1 / (\<Prod>j::nat=0..i. a j)) (nhds putnam_2011_a2_solution) at_top" sorry end

putnam_2011_a3

abbrev putnam_2011_a3_solution : ℝ × ℝ := sorry -- (-1, 2 / Real.pi) theorem putnam_2011_a3 : putnam_2011_a3_solution.2 > 0 ∧ Tendsto (fun r : ℝ => (r ^ putnam_2011_a3_solution.1 * ∫ x in Set.Ioo 0 (Real.pi / 2), x ^ r * Real.sin x) / (∫ x in Set.Ioo 0 (Real.pi / 2), x ^ r * Real.cos x)) atTop (𝓝 putnam_2011_a3_solution.2) := sorry

Find a real number $c$ and a positive number $L$ for which $\lim_{r \to \infty} \frac{r^c \int_0^{\pi/2} x^r\sin x\,dx}{\int_0^{\pi/2} x^r\cos x\,dx}=L$.

Show that $(c,L)=(-1,2/\pi)$ works.

['analysis']

Section putnam_2011_a3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2011_a3_solution := (-1, 2 / PI). Theorem putnam_2011_a3: exists (c L: R), L > 0 -> Lim_seq (fun r => (Rpower (INR r) c * RInt (fun x => x ^ r * sin x) 0 PI / 2) / (RInt (fun x => x ^ r * cos x) 0 PI / 2)) = L <-> (c, L) = putnam_2011_a3_solution. Proof. Admitted. End putnam_2011_a3.

theory putnam_2011_a3 imports Complex_Main "HOL-Analysis.Set_Integral" "HOL-Analysis.Lebesgue_Measure" begin (* there may be multiple possible correct answers *) definition putnam_2011_a3_solution :: "real \<times> real" where "putnam_2011_a3_solution \<equiv> undefined" (* (-1, 2/pi) *) theorem putnam_2011_a3: shows "let (c, L) = putnam_2011_a3_solution in (L > 0 \<and> filterlim (\<lambda>r::real. (r powr c * (set_lebesgue_integral lebesgue {0<..<(pi/2)} (\<lambda>x::real. x powr r * sin x))) / (set_lebesgue_integral lebesgue {0<..<(pi/2)} (\<lambda>x::real. x powr r * cos x))) (nhds L) at_top)" sorry end

putnam_2011_a4

abbrev putnam_2011_a4_solution : Set ℕ := sorry -- {n : ℕ | Odd n} theorem putnam_2011_a4 (n : ℕ) (nmat : Prop) (npos : n > 0) (hnmat : ∃ A : Matrix (Fin n) (Fin n) ℤ, (∀ r : Fin n, Even ((A r) ⬝ᵥ (A r))) ∧ (∀ r1 r2 : Fin n, r1 ≠ r2 → Odd ((A r1) ⬝ᵥ (A r2)))) : nmat ↔ n ∈ putnam_2011_a4_solution := sorry

For which positive integers $n$ is there an $n \times n$ matrix with integer entries such that every dot product of a row with itself is even, while every dot product of two different rows is odd?

Show that the answer is $n$ odd.

['linear_algebra']

theory putnam_2011_a4 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Analysis.Infinite_Sum" begin (* uses (nat \<Rightarrow> 'n) instead of (Fin n \<Rightarrow> 'n) *) definition putnam_2011_a4_solution :: "nat set" where "putnam_2011_a4_solution \<equiv> undefined" (* {n::nat. odd n} *) theorem putnam_2011_a4: fixes n :: nat and intinner :: "int^'n \<Rightarrow> int^'n \<Rightarrow> int" and nmat :: bool assumes npos: "n > 0" and pncard: "CARD('n) = n" and hintinner2: "\<forall>u v::int^'n. intinner u v = (\<Sum>\<^sub>\<infinity>i::'n. u$i * v$i)" and hnmat: "\<exists>A::int^'n^'n. (\<forall>r::'n. even (intinner (row r A) (row r A))) \<and> (\<forall>r1 r2::'n. r1 \<noteq> r2 \<longrightarrow> odd (intinner (row r1 A) (row r2 A)))" shows "nmat \<longleftrightarrow> n \<in> putnam_2011_a4_solution" sorry end

putnam_2011_b1

theorem putnam_2011_b1 (h k : ℤ) (hkpos : h > 0 ∧ k > 0) : ∀ ε > 0, ∃ m n : ℤ, m > 0 ∧ n > 0 ∧ ε < |h * Real.sqrt m - k * Real.sqrt n| ∧ |h * Real.sqrt m - k * Real.sqrt n| < 2 * ε := sorry

Let $h$ and $k$ be positive integers. Prove that for every $\epsilon>0$, there are positive integers $m$ and $n$ such that $\epsilon<|h\sqrt{m}-k\sqrt{n}|<2\epsilon$.

['algebra']

Section putnam_2011_b1. Require Import Reals ZArith Coquelicot.Coquelicot. Theorem putnam_2011_b1: forall (h k: Z), Z.gt h 0 /\ Z.gt k 0 -> forall (ep: R), ep > 0 /\ exists (m n: Z), ep < Rabs (IZR h * sqrt (IZR m) - IZR k * sqrt (IZR n)) < 2 * ep. Proof. Admitted. End putnam_2011_b1.

theory putnam_2011_b1 imports Complex_Main begin theorem putnam_2011_b1: fixes h k :: nat assumes hkpos: "h > 0 \<and> k > 0" shows "\<forall>\<epsilon>>0. \<exists>m n::nat. m > 0 \<and> n > 0 \<and> \<epsilon> < \<bar>h * sqrt m - k * sqrt n\<bar> \<and> \<bar>h * sqrt m - k * sqrt n\<bar> < 2*\<epsilon>" sorry end

putnam_2011_b2

abbrev putnam_2011_b2_solution : Set ℕ := sorry -- {2, 5} theorem putnam_2011_b2 (S : Set (Fin 3 → ℕ)) (t : ℕ) (t7inS : Prop) (hS : S = {s : Fin 3 → ℕ | (s 0).Prime ∧ (s 1).Prime ∧ (s 2).Prime ∧ ∃ x : ℚ, (s 0) * x ^ 2 + (s 1) * x + (s 2) = 0}) (ht7inS : t7inS = ({s ∈ S | ∃ i : Fin 3, s i = t}.encard ≥ 7)) : (t.Prime ∧ t7inS) ↔ t ∈ putnam_2011_b2_solution := sorry

Let $S$ be the set of all ordered triples $(p,q,r)$ of prime numbers for which at least one rational number $x$ satisfies $px^2+qx+r=0$. Which primes appear in seven or more elements of $S$?

Show that only the primes $2$ and $5$ appear seven or more times.

['number_theory']

theory putnam_2011_b2 imports Complex_Main "HOL-Computational_Algebra.Primes" begin definition putnam_2011_b2_solution :: "nat set" where "putnam_2011_b2_solution \<equiv> undefined" (* {2, 5} *) theorem putnam_2011_b2: fixes S :: "(nat \<times> nat \<times> nat) set" and t :: nat and t7inS :: bool assumes hS: "S \<equiv> {(p::nat,q::nat,r::nat). prime p \<and> prime q \<and> prime r \<and> (\<exists>x::rat. p*x^2 + q*x + r = 0)}" and ht7inS: "t7inS \<equiv> (card {s\<in>S. fst s = t \<or> fst (snd s) = t \<or> snd (snd s) = t} \<ge> 7)" shows "(prime t \<and> t7inS) \<longleftrightarrow> t \<in> putnam_2011_b2_solution" sorry end

putnam_2011_b3

abbrev putnam_2011_b3_solution : Prop := sorry -- True theorem putnam_2011_b3 : ((∀ f g : ℝ → ℝ, g 0 ≠ 0 → ContinuousAt g 0 → DifferentiableAt ℝ (fun x ↦ f x * g x) 0 → DifferentiableAt ℝ (fun x ↦ f x / g x) 0 → (DifferentiableAt ℝ f 0)) ↔ putnam_2011_b3_solution) := sorry

Let $f$ and $g$ be (real-valued) functions defined on an open interval containing $0$, with $g$ nonzero and continuous at $0$. If $fg$ and $f/g$ are differentiable at $0$, must $f$ be differentiable at $0$?

Prove that $f$ is differentiable.

['analysis']

Section putnam_2011_b3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2011_b3_solution := True. Theorem putnam_2011_b3: forall (f g: R -> R) (a b: R), (a < 0 < b /\ forall (x: R), a < x < b -> g x > 0 /\ continuity_pt g 0 /\ ex_derive f (g 0) /\ ex_derive (fun x => f x / g x) 0) -> ex_derive f 0 <-> putnam_2011_b3_solution. Proof. Admitted. End putnam_2011_b3.

theory putnam_2011_b3 imports Complex_Main "HOL-Analysis.Derivative" begin definition putnam_2011_b3_solution :: bool where "putnam_2011_b3_solution \<equiv> undefined" (* True *) theorem putnam_2011_b3: shows "(\<forall>f g::real\<Rightarrow>real. g 0 \<noteq> 0 \<longrightarrow> continuous_on {0} g \<longrightarrow> (\<lambda>x::real. f x*g x) differentiable_on {0} \<longrightarrow> (\<lambda>x::real. f x/g x) differentiable_on {0} \<longrightarrow> f differentiable_on {0}) \<longleftrightarrow> putnam_2011_b3_solution" sorry end

putnam_2011_b5

theorem putnam_2011_b5 (a : ℕ → ℝ) (h : ∃ A : ℝ, ∀ n : ℕ, ∫ x : ℝ, ((∑ i : Finset.range n, 1 / (1 + (x - a i) ^ 2)) ^ 2) ≤ A * n) : (∃ B : ℝ, B > 0 ∧ ∀ n : ℕ, ∑' i : Finset.range n, ∑' j : Finset.range n, (1 + (a i - a j) ^ 2) ≥ B * n ^ 3) := sorry

Let $a_1, a_2, \dots$ be real numbers. Suppose that there is a constant $A$ such that for all $n$, \[ \int_{-\infty}^\infty \left( \sum_{i=1}^n \frac{1}{1 + (x-a_i)^2} \right)^2\,dx \leq An. \] Prove there is a constant $B>0$ such that for all $n$, \[ \sum_{i,j=1}^n (1 + (a_i - a_j)^2) \geq Bn^3. \]

['analysis']

Section putnam_2011_b5. Require Import Reals Coquelicot.Coquelicot. Theorem putnam_2011_b5: forall (a: nat -> R), (exists (A: R), forall (n: nat), Lim_seq (fun nInc => (RInt (fun x => (sum_n (fun i => 1 / (1 + (x - a i) ^ 2)) n)) (-1 * INR nInc) (INR nInc)) ^ 2) <= A * INR n) -> exists (B: R), B > 0 /\ forall (n: nat), sum_n (fun i => (sum_n (fun j => 1 + (a i - a j) ^ 2)) n) n >= B * INR n ^ 3. Proof. Admitted. End putnam_2011_b5.

theory putnam_2011_b5 imports Complex_Main "HOL-Analysis.Set_Integral" "HOL-Analysis.Lebesgue_Measure" begin theorem putnam_2011_b5: fixes a :: "nat \<Rightarrow> real" assumes h: "\<exists>A::real. \<forall>n::nat. (set_lebesgue_integral lebesgue \<real> (\<lambda>x::real. (\<Sum>i::nat=0..(n-1). 1 / (1 + (x-a i)^2)) ^ 2)) \<le> A*n" shows "\<exists>B::real>0. \<forall>n::nat. (\<Sum>i::nat=0..(n-1). \<Sum>j::nat=0..(n-1). 1 + (a i-a j)^2) \<ge> B*n^3" sorry end

putnam_2011_b6

theorem putnam_2011_b6 (p : ℕ) (hp : Odd p ∧ Nat.Prime p) : {n ∈ Finset.range p | ¬ p ∣ ∑ k : Finset.range p, Nat.factorial k * n^(k : ℕ)}.ncard ≥ (p + 1)/2 := sorry

Let $p$ be an odd prime. Show that for at least $(p+1)/2$ values of $n$ in $\{0,1,2,\dots,p-1\}$, \[ \sum_{k=0}^{p-1} k! n^k \qquad \mbox{is not divisible by $p$.} \]

['number_theory']

Section putnam_2011_b6. Require Import Nat List Factorial Ensembles Finite_sets Reals Znumtheory ZArith Coquelicot.Coquelicot. Open Scope nat_scope. Theorem putnam_2011_b6: forall (p: nat), prime (Z.of_nat p) /\ odd p = true -> let l := seq 0 p in exists (E: Ensemble nat), (forall (n: nat), E n -> and (le 0 n) (lt n p)) /\ cardinal nat E ((p + 1) / 2) /\ forall (n: nat), E n -> Z.to_nat (floor (sum_n (fun k => INR (fact k * n ^ k)) p)) mod p <> 0. Proof. Admitted. End putnam_2011_b6.

theory putnam_2011_b6 imports Complex_Main "HOL-Computational_Algebra.Primes" begin theorem putnam_2011_b6: fixes p :: nat assumes hp: "odd p \<and> prime p" shows "card {n::nat\<in>{0..(p-1)}. \<not>(p dvd (\<Sum>k::nat=0..(p-1). fact k * n^k))} \<ge> (p + 1)/2" sorry end

putnam_2017_a1

abbrev putnam_2017_a1_solution : Set ℤ := sorry -- {x : ℤ | x > 0 ∧ (x = 1 ∨ 5 ∣ x)} theorem putnam_2017_a1 (Q : Set (Set ℤ)) (Spos : ∀ S ∈ Q, ∀ x ∈ S, x > 0) (S2 : ∀ S ∈ Q, 2 ∈ S) (Sn : ∀ S ∈ Q, ∀ n, n ^ 2 ∈ S → n ∈ S) (Sn5 : ∀ S ∈ Q, ∀ n, n ∈ S → (n + 5) ^ 2 ∈ S) : Set.univ \ (⋂ T ∈ Q, T) = putnam_2017_a1_solution := sorry

Let $S$ be the smallest set of positive integers such that (a) $2$ is in $S$, (b) $n$ is in $S$ whenever $n^2$ is in $S$, and (c) $(n+5)^2$ is in $S$ whenever $n$ is in $S$. Which positive integers are not in $S$?.

Show that all solutions are in the set $\{x \in \mathbb{Z}\, |\, x > 0 \land (x = 1 \lor 5 \mid x)\}

['number_theory']

Section putnam_2017_a1. From mathcomp Require Import div. Definition putnam_2017_a1_solution (A: nat -> Prop) (x: nat) := x > 0 /\ (x = 1 \/ 5 %| x = true) -> ~ A x. Theorem putnam_2017_a1: exists (A: nat -> Prop), forall (B: nat -> Prop), let valid_set (A: nat -> Prop): Prop := forall (n: nat), A 2 /\ A (n*n) -> A n /\ A n -> A ((n+5)*(n+5)) in valid_set A /\ valid_set B /\ forall (n: nat), A n -> B n -> putnam_2017_a1_solution A n. Proof. Admitted. End putnam_2017_a1.

theory putnam_2017_a1 imports Complex_Main begin definition putnam_2017_a1_solution::"int set" where "putnam_2017_a1_solution \<equiv> undefined" (* { x::int . x > 0 \<and> (x = 1 \<or> 5 dvd x) } *) theorem putnam_2017_a1: shows "putnam_2017_a1_solution = (LEAST S. 2 \<in> S \<and> (\<forall>n. n^2 \<in> S \<longrightarrow> n \<in> S) \<and> (\<forall>n. n \<in> S \<longrightarrow> (n+5)^2 \<in> S) \<and> (\<forall>n \<in> S. n > 0))" sorry end

putnam_2017_a2

theorem putnam_2017_a2 (Q : ℕ → ℝ → ℝ) (hQbase : ∀ x : ℝ, Q 0 x = 1 ∧ Q 1 x = x) (hQn : ∀ n ≥ 2, ∀ x : ℝ, Q n x = ((Q (n - 1) x) ^ 2 - 1) / Q (n - 2) x) : ∀ n > 0, ∃ P : Polynomial ℝ, (∀ i : ℕ, P.coeff i = round (P.coeff i)) ∧ Q n = P.eval := sorry

Let $Q_0(x)=1$, $Q_1(x)=x$, and $Q_n(x)=\frac{(Q_{n-1}(x))^2-1}{Q_{n-2}(x)}$ for all $n \geq 2$. Show that, whenever $n$ is a positive integer, $Q_n(x)$ is equal to a polynomial with integer coefficients.

['algebra']

Section putnam_2017_a2. Require Import Nat QArith Reals. From mathcomp Require Import seq ssrnat ssrnum ssralg poly. Local Open Scope ring_scope. Theorem putnam_2017_a2: let Q := fix q (n: nat) (x: R): R := match n with | O => R1 | S O => x | S ((S n'') as n') => Rdiv (Rminus (Rmult (q n' x) (q n' x)) 1) (q n'' x) end in forall (n: nat), ge n 0 -> exists (R: numDomainType) (p: {poly R}) (i: nat), (exists (z: Z), p`_i = if (Z.ltb z 0) then -(Z.to_nat z)%:R else (Z.to_nat z)%:R) /\ exists (z: Z), forall (x: RbaseSymbolsImpl.R), Q n x = IZR z /\ (if (Z.ltb z 0) then -(Z.to_nat z)%:R else (Z.to_nat z)%:R) = p.[n%:R]. Proof. Admitted. End putnam_2017_a2.

theory putnam_2017_a2 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2017_a2: fixes Q :: "nat \<Rightarrow> real \<Rightarrow> real" assumes hQbase: "\<forall>x::real. Q 0 x = 1 \<and> Q 1 x = x" and hQn: "\<forall>n::nat>2. \<forall>x::real. Q n x = ((Q (n-1) x)^2 - 1) / Q (n-2) x" shows "\<forall>n::nat>0. \<exists>P::real poly. (\<forall>i::nat. coeff P i = round (coeff P i)) \<and> Q n = poly P" sorry end

putnam_2017_a3

theorem putnam_2017_a3 (a b : ℝ) (f g : ℝ → ℝ) (I : ℕ → ℝ) (altb : a < b) (fgcont : ContinuousOn f (Set.Icc a b) ∧ ContinuousOn g (Set.Icc a b)) (fgimg : f '' (Set.Icc a b) ⊆ Set.Ioi 0 ∧ g '' (Set.Icc a b) ⊆ Set.Ioi 0) (fgint : ∫ x in Set.Ioo a b, f x = ∫ x in Set.Ioo a b, g x) (fneg : ∃ x : Set.Icc a b, f x ≠ g x) (hI : ∀ n > 0, I n = ∫ x in Set.Ioo a b, ((f x) ^ (n + 1)) / ((g x) ^ n)) : (∀ n > 0, I (n + 1) > I n) ∧ Tendsto I atTop atTop := sorry

Let $a$ and $b$ be real numbers with $a<b$, and let $f$ and $g$ be continuous functions from $[a,b]$ to $(0,\infty)$ such that $\int_a^b f(x)\,dx=\int_a^b g(x)\,dx$ but $f \neq g$. For every positive integer $n$, define $I_n=\int_a^b \frac{(f(x))^{n+1}}{(g(x))^n}\,dx$. Show that $I_1,I_2,I_3,\dots$ is an increasing sequence with $\lim_{n \to \infty} I_n=\infty$.

['analysis']

Section putnam_2017_a3. Require Import Reals. From Coquelicot Require Import Coquelicot Continuity RInt. Open Scope R. Theorem putnam_2017_a3: forall (a b: R), a < b -> forall (f g: R -> R) (x: R), a <= x <= b -> continuity_pt f x /\ f x > 0 /\ g x > 0 -> (RInt f a b = RInt g a b) /\ (exists (x: R), a <= x <= b /\ f x <> g x) -> let I_n (n: nat) := RInt (fun x => (f x)^(n+1)/(g x)^n) a b in forall (n: nat), I_n n < I_n (Nat.add n 1) /\ ~ ex_lim_seq I_n. Proof. Admitted. End putnam_2017_a3.

theory putnam_2017_a3 imports Complex_Main "HOL-Analysis.Set_Integral" "HOL-Analysis.Lebesgue_Measure" begin (* uses (real \<Rightarrow> real) instead of ({a..b} \<Rightarrow> {0<..}) *) theorem putnam_2017_a3: fixes a b :: real and f g :: "real \<Rightarrow> real" and I :: "nat \<Rightarrow> real" assumes altb: "a < b" and fgcont: "continuous_on {a..b} f \<and> continuous_on {a..b} g" and fgimg: "f ` {a..b} \<subseteq> {0<..} \<and> g ` {a..b} \<subseteq> {0<..}" and fgint: "set_lebesgue_integral lebesgue {a<..<b} f = set_lebesgue_integral lebesgue {a<..<b} g" and fneg: "\<exists>x::real\<in>{a..b}. f x \<noteq> g x" and hI: "\<forall>n::nat>0. I n = set_lebesgue_integral lebesgue {a<..<b} (\<lambda>x::real. (f x)^(n+1) / (g x)^n)" shows "(\<forall>n::nat>0. I (n+1) > I n) \<and> filterlim I at_top at_top" sorry end

putnam_2017_a4

theorem putnam_2017_a4 (N : ℕ) (score : Fin (2 * N) → Fin 11) (hsurj : ∀ k : Fin 11, ∃ i : Fin (2 * N), score i = k) (havg : (∑ i : Fin (2 * N), (score i : ℝ)) / (2 * N) = 7.4) : (∃ s : Finset (Fin (2 * N)), s.card = N ∧ (∑ i in s, (score i : ℝ)) / N = 7.4 ∧ (∑ i in sᶜ, (score i : ℝ)) / N = 7.4) := sorry

A class with $2N$ students took a quiz, on which the possible scores were $0,1,\dots,10$. Each of these scores occurred at least once, and the average score was exactly $7.4$. Show that the class can be divided into two groups of $N$ students in such a way that the average score for each group was exactly $7.4$.

['algebra']

Section putnam_2017_a4. Require Import Nat Ensembles List Finite_sets Reals Coquelicot.Coquelicot. From mathcomp Require Import div fintype seq ssrbool. Theorem putnam_2017_a4 (N : nat) (M : nat := mul 2 N) (score : nat -> 'I_11) (hsurj : forall (k: 'I_11), exists (i : 'I_M), score i = k) (havg : (sum_n (fun i => INR (nat_of_ord (score i))) M) / INR M = 7.4) (E_bool : nat -> bool) : exists (presS: nat -> Prop) (E: Ensemble nat), cardinal nat E N /\ forall (n: nat), E n <-> (presS n /\ and (le 0 n) (le n 10)) /\ (forall i, E_bool i = true <-> E i) /\ (sum_n (fun i => if (E_bool i) then INR (nat_of_ord (score i)) else 0) N) / INR N = 7.4 /\ (sum_n (fun i => if E_bool i then 0 else INR (nat_of_ord (score i))) N) / INR N = 7.4. Proof. Admitted. End putnam_2017_a4.

theory putnam_2017_a4 imports Complex_Main begin theorem putnam_2017_a4: fixes N :: nat and score :: "nat \<Rightarrow> nat" assumes hrng : "\<forall> i \<in> {0 .. 2 * N - 1}. score i \<in> {0 .. 10}" and hsurj: "\<forall> k \<in> {0 .. 10}. \<exists> i \<in> {0 .. 2 * N - 1}. score i = k" and havg: "(\<Sum> i = 0 .. (2 * N - 1). score i) / (2 * N) = 7.4" shows "\<exists> s \<subseteq> {0 .. 2 * N - 1}. card s = N \<and> (\<Sum> i \<in> s. score i) / N = 7.4 \<and> (\<Sum> i \<in> {0 .. 2 * N - 1} - s. score i) / N = 7.4" sorry end

putnam_2017_b1

theorem putnam_2017_b1 (lines : Set (Set (Fin 2 → ℝ)) := {L : Set (Fin 2 → ℝ) | ∃ v w : Fin 2 → ℝ, w ≠ 0 ∧ L = {p : Fin 2 → ℝ | ∃ t : ℝ, p = v + t • w}}) (L1 L2 : Set (Fin 2 → ℝ)) (L1L2lines : L1 ∈ lines ∧ L2 ∈ lines) (L1L2distinct : L1 ≠ L2) : L1 ∩ L2 ≠ ∅ ↔ (∀ lambda : ℝ, lambda ≠ 0 → ∀ P : Fin 2 → ℝ, (P ∉ L1 ∧ P ∉ L2) → ∃ A1 A2 : Fin 2 → ℝ, A1 ∈ L1 ∧ A2 ∈ L2 ∧ (A2 - P = lambda • (A1 - P))) := sorry

Let $L_1$ and $L_2$ be distinct lines in the plane. Prove that $L_1$ and $L_2$ intersect if and only if, for every real number $\lambda \neq 0$ and every point $P$ not on $L_1$ or $L_2$, there exist points $A_1$ on $L_1$ and $A_2$ on $L_2$ such that $\overrightarrow{PA_2}=\lambda \overrightarrow{PA_1}$.

['geometry']

theory putnam_2017_b1 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin theorem putnam_2017_b1: fixes lines :: "(real^2) set set" and L1 L2 :: "(real^2) set" defines "lines \<equiv> {L :: (real^2) set. \<exists> v w :: real^2. w \<noteq> 0 \<and> L = {p :: real^2. \<exists> t :: real. p = v + t *s w}}" assumes L1L2lines: "L1 \<in> lines \<and> L2 \<in> lines" and L1L2distinct: "L1 \<noteq> L2" shows "(L1 \<inter> L2 \<noteq> {}) \<longleftrightarrow> (\<forall> (lambda :: real) \<noteq> 0. \<forall> P :: real^2. (P \<notin> L1 \<and> P \<notin> L2) \<longrightarrow> (\<exists> A1 A2 :: real^2. A1 \<in> L1 \<and> A2 \<in> L2 \<and> A2 - P = lambda *s (A1 - P)))" sorry end

putnam_2017_b2

abbrev putnam_2017_b2_solution : ℕ := sorry -- 16 theorem putnam_2017_b2 (mina : ℤ) (hmina : mina ≥ 0) (S : ℤ → ℕ → ℤ := fun a k ↦ ∑ i : Fin k, a + i) (p : ℤ → ℕ → Prop := fun N k ↦ ∃ a > 0, S a k = N) (q : ℤ → Prop := fun N ↦ p N 2017 ∧ ∀ k : ℕ, k > 1 → k ≠ 2017 → ¬p N k) (hqmina : q (S mina 2017)) (hminalb : ∀ a > 0, q (S a 2017) → mina ≤ a) : (mina = putnam_2017_b2_solution) := sorry

Suppose that a positive integer $N$ can be expressed as the sum of $k$ consecutive positive integers \[ N = a + (a+1) +(a+2) + \cdots + (a+k-1) \] for $k=2017$ but for no other values of $k>1$. Considering all positive integers $N$ with this property, what is the smallest positive integer $a$ that occurs in any of these expressions?

Prove that the smallest value of $a$ is $16$.

['algebra']

Section putnam_2017_b2. Require Import Nat ZArith Coquelicot.Coquelicot. Definition putnam_2017_b2_solution := 16. Theorem putnam_2017_b2 (mina : nat) (posMin : mina > 0) (A : nat -> nat -> nat := fun a k => Z.to_nat (floor (sum_n (fun i => Raxioms.INR (a + (i + 1))) k))) (p : nat -> nat -> Prop := fun N k => exists (a: nat), a > 0 /\ A a k = N) (q : nat -> Prop := fun N => p N 2017 /\ forall (k: nat), k > 1 -> k <> 2017 -> ~ p N k) (hmina : q (A mina 2017)) (hminalb : (forall (a: nat), a > 0 /\ q (A a 2017) -> mina <= a)) : mina = putnam_2017_b2_solution. Proof. Admitted. End putnam_2017_b2.

theory putnam_2017_b2 imports Complex_Main begin definition putnam_2017_b2_solution :: nat where "putnam_2017_b2_solution \<equiv> undefined" (* 16 *) theorem putnam_2017_b2: fixes S :: "nat \<Rightarrow> nat \<Rightarrow> nat" and p :: "nat \<Rightarrow> nat \<Rightarrow> bool" and q :: "nat \<Rightarrow> bool" defines "S \<equiv> \<lambda> a k. \<Sum> i \<in> {0 .. k - 1}. a + i" and "p \<equiv> \<lambda> N k. \<exists> a > 0. S a k = N" and "q \<equiv> \<lambda> N. p N 2017 \<and> (\<forall> k > 1. k \<noteq> 2017 \<longrightarrow> \<not>p N k)" shows "(LEAST a. a > 0 \<and> q (S a 2017)) = putnam_2017_b2_solution" sorry end

putnam_2017_b3

theorem putnam_2017_b3 (f : ℝ → ℝ) (c : ℕ → ℝ) (hc : ∀ n, c n = 0 ∨ c n = 1) (hf : ∀ x, f x = ∑' n : ℕ, (c n) * x^n) : f (2/3) = 3/2 → Irrational (f 1/2) := sorry

Suppose that $f(x) = \sum_{i=0}^\infty c_i x^i$ is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational.

['number_theory']

theory putnam_2017_b3 imports Complex_Main begin theorem putnam_2017_b3: fixes f :: "real \<Rightarrow> real" and c :: "nat \<Rightarrow> real" defines hf: "f \<equiv> \<lambda>x. \<Sum>i::nat. c (i) * x^i" assumes hc: "\<forall>n. c n = 0 \<or> c n = 1" and "f (2/3) = 3/2" shows "f (1/2) \<notin> \<rat>" sorry end

putnam_2017_b4

abbrev putnam_2017_b4_solution : ℝ := sorry -- (log 2) ^ 2 theorem putnam_2017_b4 : (∑' k : ℕ, 3 * log (4 * k + 2) / (4 * k + 2) - log (4 * k + 3) / (4 * k + 3) - log (4 * k + 4) / (4 * k + 4) - log (4 * k + 5) / (4 * k + 5) = putnam_2017_b4_solution) := sorry

Evaluate the sum \begin{gather*} \sum_{k=0}^\infty \left( 3 \cdot \frac{\ln(4k+2)}{4k+2} - \frac{\ln(4k+3)}{4k+3} - \frac{\ln(4k+4)}{4k+4} - \frac{\ln(4k+5)}{4k+5} ight) \ = 3 \cdot \frac{\ln 2}{2} - \frac{\ln 3}{3} - \frac{\ln 4}{4} - \frac{\ln 5}{5} + 3 \cdot \frac{\ln 6}{6} - \frac{\ln 7}{7} \ - \frac{\ln 8}{8} - \frac{\ln 9}{9} + 3 \cdot \frac{\ln 10}{10} - \cdots . \end{gather*} (As usual, $\ln x$ denotes the natural logarithm of $x$.)

Prove that the sum equals $(\ln 2)^2$.

['algebra']

Section putnam_2017_b4. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2017_b4_solution := (ln 2 / ln 10) ^ 2. Theorem putnam_2017_b4 : Lim_seq (fun n => sum_n (fun k => let k := INR k in (3 * ln (4 * k + 2) / (4 * k + 2) - ln (4 * k + 3) / (4 * k + 3) - ln (4 * k + 4) / (4 * k + 4) - ln (4 * k + 5) / (4 * k + 5))) n) = putnam_2017_b4_solution . Proof. Admitted. End putnam_2017_b4.

theory putnam_2017_b4 imports Complex_Main begin definition putnam_2017_b4_solution :: real where "putnam_2017_b4_solution \<equiv> undefined" (* (ln 2) ^ 2 *) theorem putnam_2017_b4: shows "(\<Sum> k :: nat. 3 * ln (4 * k + 2) / (4 * k + 2) - ln (4 * k + 3) / (4 * k + 3) - ln (4 * k + 4) / (4 * k - 4) - ln (4 * k + 5) / (4 * k + 5)) = putnam_2017_b4_solution" sorry end

putnam_2017_b6

abbrev putnam_2017_b6_solution : ℕ := sorry -- 2016! / 1953! - 63! * 2016 theorem putnam_2017_b6 (S : Finset (Finset.range 64 → Finset.Icc 1 2017)) (hs : ∀ x : (Finset.range 64 → Finset.Icc 1 2017), x ∈ S ↔ (Injective x ∧ (2017 ∣ (∑ i : Finset.range 64, if i ≤ (⟨1, by norm_num⟩ : Finset.range 64) then (x i : ℤ) else i * (x i : ℤ))))) : (S.card = putnam_2017_b6_solution) := sorry

Find the number of ordered $64$-tuples $(x_0,x_1,\dots,x_{63})$ such that $x_0,x_1,\dots,x_{63}$ are distinct elements of $\{1,2,\dots,2017\}$ and \[ x_0 + x_1 + 2x_2 + 3x_3 + \cdots + 63 x_{63} \] is divisible by 2017.

Prove that the answer is $\frac{2016!}{1953!} - 63! \cdot 2016$

['algebra', 'number_theory']

theory putnam_2017_b6 imports Complex_Main begin definition putnam_2017_b6_solution::nat where "putnam_2017_b6_solution \<equiv> undefined" (* (fact 2016) / (fact 1953) - (fact 63) * 2016 *) theorem putnam_2017_b6: fixes S::"(nat list) set" defines "S \<equiv> {x. (size x = 64) \<and> (\<forall>i\<in>{0..63}. x!i \<ge> 1 \<and> x!i \<le> 2017) \<and> (\<forall>i\<in>{0..63}. \<forall>j\<in>{0..63}. x!i = x!j \<longrightarrow> i = j) \<and> (2017 dvd (x!0 + (\<Sum>i=1..63. (i * x!i))))}" shows "card S = putnam_2017_b6_solution" sorry end

putnam_2014_a1

theorem putnam_2014_a1 (f : ℝ → ℝ) (hf : ∀ x : ℝ, f x = (1 - x + x ^ 2) * Real.exp x) (hfdiff : ContDiff ℝ ⊤ f) (c : ℕ → ℝ) (hc : ∀ k : ℕ, c k = taylorCoeffWithin f k Set.univ 0) : ∀ k : ℕ, c k ≠ 0 → ∃ q : ℚ, c k = q ∧ (q.num = 1 ∨ Prime q.num.natAbs) := sorry

Prove that every nonzero coefficient of the Taylor series of \[(1 - x + x^2)e^x\] about $x=0$ is a rational number whose numerator (in lowest terms) is either $1$ or a prime number.

['analysis', 'number_theory']

theory putnam_2014_a1 imports Complex_Main "HOL-Analysis.Derivative" "HOL-Computational_Algebra.Primes" begin theorem putnam_2014_a1: fixes f::"real\<Rightarrow>real" and coeff::"nat\<Rightarrow>real" and n::nat defines "f \<equiv> \<lambda>x. (1 - x + x^2) * exp (x)" and "coeff \<equiv> \<lambda>n. (deriv^^n) f 0 / fact n" assumes non_zero : "coeff n \<noteq> 0" shows "\<exists>r::rat. coeff n = r \<and> (fst (quotient_of r) = 1 \<or> prime (fst (quotient_of r)))" sorry end

putnam_2014_a2

abbrev putnam_2014_a2_solution : ℕ → ℝ := sorry -- (fun n : ℕ => (-1) ^ (n - 1) / ((n - 1)! * (n)!)) theorem putnam_2014_a2 (n : ℕ) (A : Matrix (Fin n) (Fin n) ℝ) (npos : n > 0) (hA : ∀ i j : Fin n, A i j = 1 / min (i.1 + 1 : ℚ) (j.1 + 1)) : A.det = putnam_2014_a2_solution n := sorry

Let $A$ be the $n \times n$ matrix whose entry in the $i$-th row and $j$-th column is $\frac{1}{\min(i,j)}$ for $1 \leq i,j \leq n$. Compute $\det(A)$.

Show that the determinant is $\frac{(-1)^{n-1}}{(n-1)!n!}$.

['linear_algebra']

Section putnam_2014_a2. From mathcomp Require Import div. Theorem putnam_2014_a2: True. Proof. Admitted. End putnam_2014_a2.

theory putnam_2014_a2 imports Complex_Main "HOL-Analysis.Determinants" begin definition putnam_2014_a2_solution :: "nat \<Rightarrow> real" where "putnam_2014_a2_solution \<equiv> undefined" (* (\<lambda>n::nat. (-1)^(n-1) / (fact (n-1) * fact n)) *) theorem putnam_2014_a2: fixes n :: nat and A :: "real^'n^'n" assumes npos: "n > 0" and pncard: "CARD('n) = n" and hA: "\<exists>pnind::nat\<Rightarrow>'n. pnind ` {0..(n-1)} = UNIV \<and> (\<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. A$(pnind i)$(pnind j) = 1 / min (i+1) (j+1))" shows "det A = putnam_2014_a2_solution n" sorry end

putnam_2014_a3

abbrev putnam_2014_a3_solution : ℝ := sorry -- 3 / 7 theorem putnam_2014_a3 (a : ℕ → ℝ) (a0 : a 0 = 5 / 2) (ak : ∀ k ≥ 1, a k = (a (k - 1)) ^ 2 - 2) : Tendsto (fun n : ℕ => ∏ k in Finset.range n, (1 - 1 / a k)) atTop (𝓝 putnam_2014_a3_solution) := sorry

Let \( a_0 = \frac{5}{2} \) and \( a_k = a_{k-1}^2 - 2 \) for \( k \geq 1 \). Compute \( \prod_{k=0}^{\infty} \left(1 - rac{1}{a_k}\right) \) in closed form.

Show that the solution is 3/7.

['number_theory']

Section putnam_2014_a3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2014_a3_solution := 3 / 7. Theorem putnam_2014_a3: let fix prod_n (m: nat -> R) (n : nat) : R := match n with | O => m 0%nat | S n' => m n' * prod_n m n' end in let fix a (n: nat) := match n with | O => 5 / 2 | S n' => (a n') ^ 2 - 2 end in Lim_seq (fun n => prod_n (fun k => 1 - 1 / a k) n) = putnam_2014_a3_solution. Proof. Admitted. End putnam_2014_a3.

theory putnam_2014_a3 imports Complex_Main begin definition putnam_2014_a3_solution :: real where "putnam_2014_a3_solution \<equiv> undefined" (* 3/7 *) theorem putnam_2014_a3: fixes a :: "nat \<Rightarrow> real" assumes a0: "a 0 = 5/2" and ak: "\<forall>k::nat\<ge>1. a k = (a (k-1))^2 - 2" shows "filterlim (\<lambda>n::nat. (\<Prod>k::nat=0..(n-1). (1 - 1/(a k)))) (nhds putnam_2014_a3_solution) at_top" sorry end

putnam_2014_a5

theorem putnam_2014_a5 (P : ℕ → Polynomial ℂ) (hP : ∀ n, P n = ∑ i in Finset.Icc 1 n, i * Polynomial.X ^ (i - 1)) : ∀ (j k : ℕ), (j > 0 ∧ k > 0) → j ≠ k → IsCoprime (P j) (P k) := sorry

Let \[ P_n(x) = 1 + 2 x + 3 x^2 + \cdots + n x^{n-1}.\] Prove that the polynomials $P_j(x)$ and $P_k(x)$ are relatively prime for all positive integers $j$ and $k$ with $j \neq k$.

['algebra']

theory putnam_2014_a5 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2014_a5: fixes pf :: "nat \<Rightarrow> real poly" and j k :: nat assumes hdeg : "\<forall>n. degree (pf n) = n - 1" and hpf : "\<forall>n. \<forall>p<n. coeff (pf n) p = p + 1" and hjk : "j \<noteq> k" and jpos : "j > 0" and kpos : "k > 0" shows "\<exists> a b:: real poly. a * (pf j) + b * (pf k) = 1" sorry end

putnam_2014_a6

abbrev putnam_2014_a6_solution : ℕ → ℕ := sorry -- (fun n : ℕ => n ^ n) theorem putnam_2014_a6 (n : ℕ) (kex : ℕ → Prop) (npos : n > 0) (hkex : ∀ k ≥ 1, kex k = ∃ M N : Fin k → Matrix (Fin n) (Fin n) ℝ, ∀ i j : Fin k, ((∃ p : Fin n, (M i * N j) p p = 0) ↔ i ≠ j)) : (putnam_2014_a6_solution n ≥ 1 ∧ kex (putnam_2014_a6_solution n)) ∧ (∀ k ≥ 1, kex k → k ≤ putnam_2014_a6_solution n) := sorry

Let \( n \) be a positive integer. What is the largest \( k \) for which there exist \( n \times n \) matrices \( M_1, \ldots, M_k \) and \( N_1, \ldots, N_k \) with real entries such that for all \( i \) and \( j \), the matrix product \( M_i N_j \) has a zero entry somewhere on its diagonal if and only if \( i \neq j \)?

Show that the solution has the form k \<= n ^ n.

['linear_algebra']

theory putnam_2014_a6 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin (* uses (nat \<Rightarrow> real^'n^'n) instead of (Fin k \<Rightarrow> real^'n^'n) *) definition putnam_2014_a6_solution :: "nat \<Rightarrow> nat" where "putnam_2014_a6_solution \<equiv> undefined" (* (\<lambda>n::nat. n^n) *) theorem putnam_2014_a6: fixes n :: nat and kex :: "nat \<Rightarrow> bool" assumes npos: "n > 0" and pncard: "CARD('n) = n" and hkex: "\<forall>k::nat\<ge>1. kex k = (\<exists>M N::nat\<Rightarrow>real^'n^'n. \<forall>i::nat\<in>{0..(k-1)}. \<forall>j::nat\<in>{0..(k-1)}. ((\<exists>p::'n. (M i ** N j)$p$p = 0) \<longleftrightarrow> i \<noteq> j))" shows "(GREATEST k::nat. k \<ge> 1 \<and> kex k) = putnam_2014_a6_solution n" sorry end

putnam_2014_b1

abbrev putnam_2014_b1_solution : Set ℕ := sorry -- {n : ℕ | n > 0 ∧ ¬∃ a ∈ digits 10 n, a = 0} theorem putnam_2014_b1 (overexpansion : ℕ → List ℕ → Prop := fun N d ↦ N = ∑ i : Fin d.length, (d.get i) * 10 ^ i.1 ∧ d.getLastI ≠ 0 ∧ ∀ a ∈ d, a ∈ Finset.range 11) (S : Set ℕ) (hS : ∀ N : ℕ, N ∈ S ↔ N > 0 ∧ ∃! d : List ℕ, overexpansion N d) : (S = putnam_2014_b1_solution) := sorry

A \emph{base $10$ over-expansion} of a positive integer $N$ is an expression of the form \[ N = d_k 10^k + d_{k-1} 10^{k-1} + \cdots + d_0 10^0 \] with $d_k \neq 0$ and $d_i \in \{0,1,2,\dots,10\}$ for all $i$. For instance, the integer $N = 10$ has two base $10$ over-expansions: $10 = 10 \cdot 10^0$ and the usual base $10$ expansion $10 = 1 \cdot 10^1 + 0 \cdot 10^0$. Which positive integers have a unique base $10$ over-expansion?

Prove that the answer is the integers with no $0$'s in their usual base $10$ expansion.

['algebra']

Section putnam_2014_b1. Require Import Nat Ensembles List. From mathcomp Require Import div fintype seq ssrbool. Fixpoint hd (n: nat) (l:list 'I_n) := match l with | nil => 0 | x :: _ => x end. Fixpoint expandl (n: nat) (l : list 'I_n) : nat := match l with | nil => 0 | h :: t => (nat_of_ord h) * 10 ^ (length l - 1) + (expandl n t) end. Definition putnam_2014_b1_solution : Ensemble nat := fun n => n > 0 /\ exists l : list 'I_10, 0 <> hd 10 l /\ In ord0 l /\ expandl 10 l = n. Theorem putnam_2014_b1 (n := 11) (overexpansion : nat -> list 'I_n -> Prop := fun N l => N = expandl 11 l) (A : Ensemble nat) (hA : forall N : nat, A N <-> N > 0 /\ exists! (l: list 'I_n), overexpansion N l) : (A = putnam_2014_b1_solution). Proof. Admitted. End putnam_2014_b1.

theory putnam_2014_b1 imports Complex_Main begin definition putnam_2014_b1_solution :: "nat set" where "putnam_2014_b1_solution \<equiv> undefined" (* {n::nat. n > 0 \<and> (\<forall>d::nat. (\<lfloor>n / 10^d\<rfloor> > 0) \<longrightarrow> (\<lfloor>n / 10^d\<rfloor> mod 10 \<noteq> 0))} *) theorem putnam_2014_b1: fixes overexpansion :: "nat \<Rightarrow> (nat list) \<Rightarrow> bool" and S :: "nat set" defines "overexpansion \<equiv> (\<lambda>(N::nat)(d::nat list). N = (\<Sum>i::nat=0..((length d)-1). (d!i)*10^i) \<and> last d \<noteq> 0 \<and> (\<forall>i::nat\<in>{0..((length d)-1)}. d!i \<in> {0..10}))" assumes hS: "\<forall>N::nat. (N \<in> S \<longleftrightarrow> (N > 0 \<and> (\<exists>!d::nat list. overexpansion N d)))" shows "S = putnam_2014_b1_solution" sorry end

putnam_2014_b2

abbrev putnam_2014_b2_solution : ℝ := sorry -- Real.log (4 / 3) theorem putnam_2014_b2 (fbound : (ℝ → ℝ) → Prop := (fun f : ℝ → ℝ => ∀ x : Set.Icc (1 : ℝ) 3, -1 ≤ f x ∧ f x ≤ 1)) (finteq0 : (ℝ → ℝ) → Prop := (fun f : ℝ → ℝ => (∫ x in Set.Ioo 1 3, f x) = 0)) (fint : (ℝ → ℝ) → ℝ := (fun f : ℝ → ℝ => ∫ x in Set.Ioo 1 3, (f x) / x)) : (∃ f : ℝ → ℝ, fbound f ∧ finteq0 f ∧ fint f = putnam_2014_b2_solution) ∧ (∀ f : ℝ → ℝ, (fbound f ∧ finteq0 f) → fint f ≤ putnam_2014_b2_solution) := sorry

Suppose that \( f \) is a function on the interval \([1,3]\) such that \(-1 \leq f(x) \leq 1\) for all \( x \) and \( \int_{1}^{3} f(x) \, dx = 0 \). How large can \(\int_{1}^{3} \frac{f(x)}{x} \, dx \) be?

Show that the solution is log (4 / 3).

['analysis']

Section putnam_2014_b2. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2014_b2_solution := ln (4 / 3) / ln 10. Theorem putnam_2014_b2: exists (m: R), ((forall (f: R -> R) (x: R), 1 <= x <= 3 /\ -1 <= f x <= 1 /\ RInt f 0 3 = 0 -> RInt (fun x => f x / x) 1 3 <= m) /\ (exists (f: R -> R) (x: R), 1 <= x <= 3 /\ -1 <= f x <= 1 /\ RInt f 0 3 = 0 -> RInt (fun x => f x / x) 1 3 = m)) <-> m = putnam_2014_b2_solution. Proof. Admitted. End putnam_2014_b2.

theory putnam_2014_b2 imports Complex_Main "HOL-Analysis.Interval_Integral" begin (* uses (real \<Rightarrow> real) instead of ({1..3} \<Rightarrow> real) *) definition putnam_2014_b2_solution :: real where "putnam_2014_b2_solution \<equiv> undefined" (* ln (4/3) *) theorem putnam_2014_b2: fixes fbound :: "(real \<Rightarrow> real) \<Rightarrow> bool" and finteq0 :: "(real \<Rightarrow> real) \<Rightarrow> bool" and fint :: "(real \<Rightarrow> real) \<Rightarrow> real" defines "fbound \<equiv> (\<lambda>f::real\<Rightarrow>real. (\<forall>x::real\<in>{1..3}. -1 \<le> f x \<and> f x \<le> 1))" and "finteq0 \<equiv> (\<lambda>f::real\<Rightarrow>real. (interval_lebesgue_integral lebesgue 1 3 f) = 0)" and "fint \<equiv> (\<lambda>f::real\<Rightarrow>real. interval_lebesgue_integral lebesgue 1 3 (\<lambda>x::real. (f x)/x))" shows "(GREATEST ffint::real. (\<exists>f::real\<Rightarrow>real. fbound f \<and> finteq0 f \<and> ffint = fint f)) = putnam_2014_b2_solution" sorry end

putnam_2014_b3

theorem putnam_2014_b3 (m n : ℕ) (A : Matrix (Fin m) (Fin n) ℚ) (mnpos : m > 0 ∧ n > 0) (Aprime : {p : ℕ | p.Prime ∧ ∃ (i : Fin m) (j : Fin n), |A i j| = p}.encard ≥ m + n) : A.rank ≥ 2 := sorry

Let $A$ be an $m \times n$ matrix with rational entries. Suppose that there are at least $m+n$ distinct prime numbers among the absolute values of the entries of $A$. Show that the rank of $A$ is at least 2.

['linear_algebra', 'number_theory']

theory putnam_2014_b3 imports Complex_Main "HOL-Analysis.Cartesian_Space" "HOL-Computational_Algebra.Primes" begin theorem putnam_2014_b3: fixes m n :: nat and A :: "rat^'n^'m" assumes mnpos: "m > 0 \<and> n > 0" and pmpncard: "CARD('m) = m \<and> CARD('n) = n" and Aprime: "card {p::nat. prime p \<and> (\<exists>(i::'m)(j::'n). \<bar>A$i$j\<bar> = p)} \<ge> m + n" shows "rank A \<ge> 2" sorry end

putnam_2014_b4

theorem putnam_2014_b4 (n : ℕ) (P: Polynomial ℂ) (npos : n > 0) (Px : P.degree = n ∧ ∀ k ∈ Set.Icc 0 n, P.coeff k = 2 ^ (k * (n - k))) : ∀ r ∈ P.roots, r.im = 0 := sorry

Show that for each positive integer \( n \), all the roots of the polynomial $ \sum_{k=0}^{n} 2^k(n-k)x^k $ are real numbers.

['algebra']

Section putnam_2014_b4. Require Import Reals Coquelicot.Coquelicot. Open Scope C. Theorem putnam_2014_b4: let fix sum_n_C (m: nat -> C) (n : nat) : C := match n with | O => m 0%nat | S n' => m n' + sum_n_C m n' end in forall (n: nat) (r: C), sum_n_C (fun k => (2 * ((INR k) * (INR n - INR k)) * r ^ k)) n = 0 <-> exists (q: R), r = RtoC q. Proof. Admitted. End putnam_2014_b4.

theory putnam_2014_b4 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin theorem putnam_2014_b4: fixes n :: nat and P :: "complex poly" assumes npos: "n > 0" and Px: "degree P = n \<and> (\<forall>k::nat\<in>{0..n}. coeff P k = 2 ^ (k*(n-k)))" shows "\<forall>r::complex. (poly P r = 0 \<longrightarrow> Im r = 0)" sorry end

putnam_2014_b6

theorem putnam_2014_b6 (f : ℝ → ℝ) (hlip : ∃ K > 0, ∀ x ∈ Icc 0 1, ∀ y ∈ Icc 0 1, |f x - f y| ≤ K * |x - y|) (hrat : ∀ r ∈ Icc (0 : ℚ) 1, ∃ a b : ℤ, f r = a + b * r) : (∃ I : Finset (Interval ℝ), (∀ Ii ∈ I, ∃ m c : ℝ, ∀ x ∈ coeHom Ii, f x = c + m * x) ∧ Icc 0 1 = ⋃ Ii ∈ I, coeHom Ii) := sorry

Let $f: [0,1] \to \mathbb{R}$ be a function for which there exists a constant $K>0$ such that $\left| f(x) - f(y) \right| \leq K \left| x - y \right|$ for all $x,y \in [0,1]$. Suppose also that for each rational number $r \in [0,1]$, there exist integers $a$ and $b$ such that $f(r) = a + br$. Prove that there exist finitely many intervals $I_1, \dots, I_n$ such that $f$ is a linear function on each $I_i$ and $[0,1] = \bigcup_{i=1}^n I_i$.

['analysis', 'number_theory']

theory putnam_2014_b6 imports Complex_Main begin (* uses (real \<Rightarrow> real) instead of ({0..1} \<Rightarrow> real) *) theorem putnam_2014_b6: fixes f :: "real \<Rightarrow> real" assumes hlip: "\<exists>K::real>0. \<forall>x::real\<in>{0..1}. \<forall>y::real\<in>{0..1}. \<bar>f x - f y\<bar> \<le> K*\<bar>x - y\<bar>" and hrat: "\<forall>r::rat\<in>{0..1}. \<exists>a b::int. f (real_of_rat r) = a + b*r" shows "\<exists>I::(real set) set. finite I \<and> (\<forall>Ii\<in>I. (\<exists>a b::real. a \<le> b \<and> Ii = {a..b}) \<and> (\<exists>m c::real. \<forall>x::real\<in>Ii. f x = c + m*x)) \<and> {0..1} = \<Union> I" sorry end

putnam_1988_a2

abbrev putnam_1988_a2_solution : Prop := sorry -- True theorem putnam_1988_a2 (f : ℝ → ℝ := fun x ↦ Real.exp (x ^ 2)) : ((∃ a b : ℝ, a < b ∧ ∃ g : ℝ → ℝ, (∃ x ∈ Ioo a b, g x ≠ 0) ∧ DifferentiableOn ℝ g (Ioo a b) ∧ ∀ x ∈ Ioo a b, deriv (fun y ↦ f y * g y) x = (deriv f x) * (deriv g x)) ↔ putnam_1988_a2_solution) := sorry

A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(fg)' = f'g'$. If $f(x)=e^{x^2}$, determine, with proof, whether there exists an open interval $(a,b)$ and a nonzero function $g$ defined on $(a,b)$ such that this wrong product rule is true for $x$ in $(a,b)$.

Show that such $(a,b)$ and $g$ exist.

['analysis']

Section putnam_1988_a2. Require Import Basics Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1988_a2_solution := True. Theorem putnam_1988_a2: let f (x: R) := exp (pow x 2) in exists (a b: R) (g: R -> R), forall (x: R), a < x < b -> Derive (compose f g) = compose (Derive f) (Derive g) <-> putnam_1988_a2_solution. Proof. Admitted. End putnam_1988_a2.

theory putnam_1988_a2 imports Complex_Main "HOL-Analysis.Derivative" begin definition putnam_1988_a2_solution :: bool where "putnam_1988_a2_solution \<equiv> undefined" (* True *) theorem putnam_1988_a2: fixes f :: "real \<Rightarrow> real" defines "f \<equiv> \<lambda> x. exp (x ^ 2)" shows "(\<exists> a b :: real. a < b \<and> (\<exists> g :: real \<Rightarrow> real. (\<exists> x \<in> {a<..<b}. g x \<noteq> 0) \<and> g differentiable_on {a<..<b} \<and> (\<forall> x \<in> {a<..<b}. deriv (\<lambda> y. f y * g y) x = (deriv f x) * (deriv g x)))) \<longleftrightarrow> putnam_1988_a2_solution" sorry end

putnam_1988_a3

abbrev putnam_1988_a3_solution : Set ℝ := sorry -- {x | x > 1 / 2} theorem putnam_1988_a3 : ({x : ℝ | ∃ L : ℝ, Tendsto (fun t ↦ ∑ n in Finset.Icc (1 : ℕ) t, (((1 / n) / Real.sin (1 / n) - 1) ^ x)) ⊤ (𝓝 L)} = putnam_1988_a3_solution) := sorry

Determine, with proof, the set of real numbers $x$ for which \[ \sum_{n=1}^\infty \left( \frac{1}{n} \csc \frac{1}{n} - 1 \right)^x \] converges.

Show that the series converges if and only if $x > \frac{1}{2}$.

['analysis']

Section putnam_1988_a3. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1988_a3_solution (a: R) := a > 1/2. Theorem putnam_1988_a3: forall (a: R), ex_lim_seq (fun m => sum_n (fun n => Rpower (1/INR n * (1 / sin (1/INR n) - 1)) a) m) <-> putnam_1988_a3_solution a. Proof. Admitted. End putnam_1988_a3.

theory putnam_1988_a3 imports Complex_Main begin definition putnam_1988_a3_solution :: "real set" where "putnam_1988_a3_solution \<equiv> undefined" (* {x :: real. x > 1 / 2} *) theorem putnam_1988_a3: shows "{x :: real. convergent (\<lambda> t :: nat. (\<Sum> n = 1..t. (1 / n) / sin (1 / n) - 1) powr x)} = putnam_1988_a3_solution" sorry end

putnam_1988_a4

abbrev putnam_1988_a4_solution : Prop × Prop := sorry -- (True, False) theorem putnam_1988_a4 (p : ℕ → Prop := fun n ↦ ∀ color : (Fin 2 → ℝ) → Fin n, ∃ p q : Fin 2 → ℝ, color p = color q ∧ Euclidean.dist p q = 1) : (let (a, b) := putnam_1988_a4_solution; (p 3 ↔ a) ∧ (p 9 ↔ b)) := sorry

\begin{enumerate} \item[(a)] If every point of the plane is painted one of three colors, do there necessarily exist two points of the same color exactly one inch apart? \item[(b)] What if ``three'' is replaced by ``nine''? \end{enumerate}

Prove that the points must exist with three colors, but not necessarily with nine.

['geometry', 'combinatorics']

theory putnam_1988_a4 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" begin definition putnam_1988_a4_solution :: "bool \<times> bool" where "putnam_1988_a4_solution \<equiv> (True, False)" (* (True, False) *) theorem putnam_1988_a4: fixes p :: "nat \<Rightarrow> bool" defines "p \<equiv> \<lambda> n :: nat. (\<forall> color :: real^2 \<Rightarrow> nat. (\<forall> v :: real^2. color v < n) \<longrightarrow> (\<exists> p q :: real^2. color p = color q \<and> dist p q = 1))" shows "let (a, b) = putnam_1988_a4_solution in (p 3 \<longleftrightarrow> a) \<and> (p 9 \<longleftrightarrow> b)" sorry end

putnam_1988_a5

theorem putnam_1988_a5 : (∃ f : ℝ → ℝ, (∀ x > 0, f (f x) = 6 * x - f x ∧ f x > 0) ∧ (∀ g : ℝ → ℝ, (∀ x > 0, g (g x) = 6 * x - g x ∧ g x > 0) → (∀ x > 0, f x = g x))) := sorry

Prove that there exists a \emph{unique} function $f$ from the set $\mathrm{R}^+$ of positive real numbers to $\mathrm{R}^+$ such that \[ f(f(x)) = 6x-f(x) \] and \[ f(x)>0 \] for all $x>0$.

['analysis']

Section putnam_1988_a5. Require Import Basics Reals. Open Scope R. Theorem putnam_1988_a5: exists! (f: R -> R), forall (x: R), (x > 0 -> f x > 0) -> (compose f f) x = 6 * x - f x. Proof. Admitted. End putnam_1988_a5.

theory putnam_1988_a5 imports Complex_Main begin theorem putnam_1988_a5: shows "\<exists> f :: real \<Rightarrow> real. (\<forall> x > 0. f (f x) = 6 * x - f x \<and> f x > 0) \<and> (\<forall> g :: real \<Rightarrow> real. (\<forall> x > 0. g (g x) = 6 * x - g x \<and> g x > 0) \<longrightarrow> (\<forall> x > 0. f x = g x))" sorry end

putnam_1988_a6

abbrev putnam_1988_a6_solution : Prop := sorry -- True theorem putnam_1988_a6 : (∀ (F V : Type*) (_ : Field F) (_ : AddCommGroup V) (_ : Module F V) (_ : FiniteDimensional F V) (n : ℕ) (A : Module.End F V) (evecs : Set V), (n = FiniteDimensional.finrank F V ∧ evecs ⊆ {v : V | ∃ f : F, A.HasEigenvector f v} ∧ evecs.encard = n + 1 ∧ (∀ sevecs : Fin n → V, (Set.range sevecs ⊆ evecs ∧ (Set.range sevecs).encard = n) → LinearIndependent F sevecs)) → (∃ c : F, A = c • LinearMap.id)) ↔ putnam_1988_a6_solution := sorry

If a linear transformation $A$ on an $n$-dimensional vector space has $n+1$ eigenvectors such that any $n$ of them are linearly independent, does it follow that $A$ is a scalar multiple of the identity? Prove your answer.

Show that the answer is yes, $A$ must be a scalar multiple of the identity.

['linear_algebra']

theory putnam_1988_a6 imports Complex_Main "HOL-Algebra.Ring" "HOL-Analysis.Finite_Cartesian_Product" begin (* This problem requires a quantifier over structures over arbitrary types, so those types are assumed to be uncountable. *) definition putnam_1988_a6_solution :: bool where "putnam_1988_a6_solution \<equiv> undefined" (* True *) theorem putnam_1988_a6: assumes pacard: "\<exists> amap :: 'a \<Rightarrow> real. surj amap" and pbcard: "\<exists> bmap :: 'b \<Rightarrow> (real^nat). surj bmap" shows "(\<forall> (F :: ('a,'m) ring_scheme) (V :: ('b,'n) ring_scheme) (scale :: 'a \<Rightarrow> 'b \<Rightarrow> 'b) (A :: 'b \<Rightarrow> 'b) (n :: nat). n > 0 \<and> field F \<and> abelian_group V \<and> (\<forall> a \<in> carrier F. \<forall> b \<in> carrier F. \<forall> x \<in> carrier V. \<forall> y \<in> carrier V. scale a (x \<oplus>\<^bsub>V\<^esub> y) = scale a x \<oplus>\<^bsub>V\<^esub> scale a y \<and> scale (a \<oplus>\<^bsub>F\<^esub> b) x = scale a x \<oplus>\<^bsub>V\<^esub> scale b x \<and> scale a (scale b x) = scale (a \<otimes>\<^bsub>F\<^esub> b) x \<and> scale \<one>\<^bsub>F\<^esub> x = x \<and> A (scale a x \<oplus>\<^bsub>V\<^esub> scale b y) = scale a (A x) \<oplus>\<^bsub>V\<^esub> scale b (A y)) \<and> n = (GREATEST k :: nat. \<exists> s :: 'b list. set s \<subseteq> carrier V \<and> length s = k \<and> card (set s) = k \<longrightarrow> (\<forall> coeffs :: 'a list. set coeffs \<subseteq> carrier F \<and> length coeffs = n \<and> (\<Oplus>\<^bsub>V\<^esub> i \<in> {0..<n}. scale (coeffs!i) (s!i)) = \<zero>\<^bsub>V\<^esub> \<longrightarrow> (\<forall> i \<in> {0..<n}. coeffs!i = \<zero>\<^bsub>F\<^esub>))) \<and> (\<exists> evecs \<subseteq> ((carrier V) - {\<zero>\<^bsub>V\<^esub>}). card evecs = n + 1 \<and> (\<forall> evec \<in> evecs. \<exists> a \<in> carrier F. A evec = scale a evec) \<and> (\<forall> sevecs :: 'b list. set sevecs \<subseteq> evecs \<and> length sevecs = n \<and> card (set sevecs) = n \<longrightarrow> (\<forall> coeffs :: 'a list. set coeffs \<subseteq> carrier F \<and> length coeffs = n \<and> (\<Oplus>\<^bsub>V\<^esub> i \<in> {0..<n}. scale (coeffs!i) (sevecs!i)) = \<zero>\<^bsub>V\<^esub> \<longrightarrow> (\<forall> i \<in> {0..<n}. coeffs!i = \<zero>\<^bsub>F\<^esub>)))) \<longrightarrow> (\<exists> a \<in> carrier F. \<forall> x \<in> carrier V. A x = scale a x)) \<longleftrightarrow> putnam_1988_a6_solution" sorry end

putnam_1988_b1

theorem putnam_1988_b1 : ∀ a ≥ 2, ∀ b ≥ 2, ∃ x y z : ℤ, x > 0 ∧ y > 0 ∧ z > 0 ∧ a * b = x * y + x * z + y * z + 1 := sorry

A \emph{composite} (positive integer) is a product $ab$ with $a$ and $b$ not necessarily distinct integers in $\{2,3,4,\dots\}$. Show that every composite is expressible as $xy+xz+yz+1$, with $x,y,z$ positive integers.

['number_theory', 'algebra']

Section putnam_1988_b1. Require Import ZArith Znumtheory. Open Scope Z. Theorem putnam_1988_b1: forall (n: Z), n > 3 /\ ~ prime n -> exists (a b c: Z), a > 0 /\ b > 0 /\ c > 0 /\ n = a * b + b * c + c * a + 1. Proof. Admitted. End putnam_1988_b1.

theory putnam_1988_b1 imports Complex_Main begin theorem putnam_1988_b1: shows "\<forall> (a :: int) \<ge> 2. \<forall> (b :: int) \<ge> 2. \<exists> x y z:: int. x > 0 \<and> y > 0 \<and> z > 0 \<and> a * b = x * y + x * z + y * z + 1" sorry end

putnam_1988_b2

abbrev putnam_1988_b2_solution : Prop := sorry -- True theorem putnam_1988_b2 : (∀ x y : ℝ, (y ≥ 0 ∧ y * (y + 1) ≤ (x + 1) ^ 2) → (y * (y - 1) ≤ x ^ 2)) ↔ putnam_1988_b2_solution := sorry

Prove or disprove: If $x$ and $y$ are real numbers with $y \geq 0$ and $y(y+1) \leq (x+1)^2$, then $y(y-1) \leq x^2$.

Show that this is true.

['algebra']

Section putnam_1988_b2. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Definition putnam_1988_b2_solution := True. Theorem putnam_1988_b2: forall (a: R), a >= 0 -> forall (x: R), pow (x + 1) 2 >= a * (a + 1) -> pow x 2 >= a * (a - 1) <-> putnam_1988_b2_solution. Proof. Admitted. End putnam_1988_b2.

theory putnam_1988_b2 imports Complex_Main begin definition putnam_1988_b2_solution :: bool where "putnam_1988_b2_solution \<equiv> undefined" (* True *) theorem putnam_1988_b2: shows "(\<forall> x y :: real. (y \<ge> 0 \<and> y * (y + 1) \<le> (x + 1) ^ 2) \<longrightarrow> y * (y - 1) \<le> x ^ 2) \<longleftrightarrow> putnam_1988_b2_solution" sorry end

putnam_1988_b3

abbrev putnam_1988_b3_solution : ℝ := sorry -- (1 + Real.sqrt 3) / 2 theorem putnam_1988_b3 (r : ℤ → ℝ) (hr : ∀ n ≥ 1, (∃ c d : ℤ, (c ≥ 0 ∧ d ≥ 0) ∧ c + d = n ∧ r n = |c - d * Real.sqrt 3|) ∧ (∀ c d : ℤ, (c ≥ 0 ∧ d ≥ 0 ∧ c + d = n) → |c - d * Real.sqrt 3| ≥ r n)) : putnam_1988_b3_solution > 0 ∧ (∀ n : ℤ, n ≥ 1 → r n ≤ putnam_1988_b3_solution) ∧ (∀ g > 0, (∀ n ≥ 1, r n ≤ g) → g ≥ putnam_1988_b3_solution) := sorry

For every $n$ in the set $N=\{1,2,\dots\}$ of positive integers, let $r_n$ be the minimum value of $|c-d \sqrt{3}|$ for all nonnegative integers $c$ and $d$ with $c+d=n$. Find, with proof, the smallest positive real number $g$ with $r_n \leq g$ for all $n \in N$.

Show that the smallest such $g$ is $(1+\sqrt{3})/2$.

['algebra']

Section putnam_1988_b3. Require Import Reals Coquelicot.Coquelicot. Definition putnam_1988_b3_solution := (1 + sqrt 3) / 2. Theorem putnam_1988_b3 (r : Z -> R) (hr : forall (n: Z), Z.ge n 1 -> (exists c d : Z, (Z.ge c 0 /\ Z.ge d 0) /\ Z.eq (Z.add c d) n /\ r n = Rabs (IZR c - IZR d * sqrt 3)) /\ (forall c d : Z, (Z.ge c 0 /\ Z.ge d 0 /\ (Z.add c d) = n) -> Rabs (IZR c - IZR d * sqrt 3) >= r n)) : putnam_1988_b3_solution > 0 /\ (forall n : Z, Z.ge n 1 -> r n <= putnam_1988_b3_solution) /\ (forall (g: R), g > 0 -> (forall (n: Z), Z.ge n 1 /\ r n <= g) -> g >= putnam_1988_b3_solution). Proof. Admitted. End putnam_1988_b3.

theory putnam_1988_b3 imports Complex_Main begin definition putnam_1988_b3_solution :: real where "putnam_1988_b3_solution \<equiv> undefined" (* (1 + sqrt 3) / 2 *) theorem putnam_1988_b3: fixes r :: "nat \<Rightarrow> real" assumes hr: "\<forall> n \<ge> 1. r n = (LEAST s. \<exists> c d :: nat. c + d = n \<and> s = \<bar>c - d * (sqrt 3)\<bar>)" shows "(LEAST g :: real. g > 0 \<and> (\<forall> n \<ge> 1. r n \<le> g)) = putnam_1988_b3_solution" sorry end

putnam_1988_b4

theorem putnam_1988_b4 (a : ℕ → ℝ) (appos : (ℕ → ℝ) → Prop) (apconv : (ℕ → ℝ) → Prop) (apposconv : (ℕ → ℝ) → Prop) (happos : ∀ a' : ℕ → ℝ, appos a' = ∀ n ≥ 1, a' n > 0) (hapconv : ∀ a' : ℕ → ℝ, apconv a' = ∃ s : ℝ, Tendsto (fun N : ℕ => ∑ n : Set.Icc 1 N, a' n) atTop (𝓝 s)) (happosconv : ∀ a' : ℕ → ℝ, apposconv a' = (appos a' ∧ apconv a')) : apposconv a → apposconv (fun n : ℕ => (a n) ^ ((n : ℝ) / (n + 1))) := sorry

Prove that if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers, then so is $\sum_{n=1}^\infty (a_n)^{n/(n+1)}$.

['analysis']

Section putnam_1988_b4. Require Import Reals Coquelicot.Coquelicot. Open Scope R. Theorem putnam_1988_b4: let b (n: nat) (a: nat -> R) := Rpower (a n) (INR n/(INR n + 1)) in forall (a: nat -> R) (n: nat), a n > 0 -> ex_lim_seq (fun n => sum_n a n) -> ex_lim_seq (fun n => sum_n (fun m => (b m a)) n). Proof. Admitted. End putnam_1988_b4.

theory putnam_1988_b4 imports Complex_Main begin theorem putnam_1988_b4: fixes a :: "nat \<Rightarrow> real" and appos :: "(nat \<Rightarrow> real) \<Rightarrow> bool" and apconv :: "(nat \<Rightarrow> real) \<Rightarrow> bool" and apposconv :: "(nat \<Rightarrow> real) \<Rightarrow> bool" defines "appos \<equiv> \<lambda> a' :: nat \<Rightarrow> real. \<forall> n \<ge> 1. a' n > 0" and "apconv \<equiv> \<lambda> a' :: nat \<Rightarrow> real. convergent (\<lambda> N :: nat. \<Sum> n = 1..N. a' n)" and "apposconv \<equiv> \<lambda> a'. appos a' \<and> apconv a'" shows "apposconv a \<longrightarrow> apposconv (\<lambda> n :: nat. (a n) powr (n / (n + 1)))" sorry end

putnam_1988_b5

abbrev putnam_1988_b5_solution : ℕ → ℕ := sorry -- (fun n : ℕ => 2 * n) theorem putnam_1988_b5 (n : ℕ) (Mn : Matrix (Fin (2 * n + 1)) (Fin (2 * n + 1)) ℝ) (npos : n > 0) (Mnskewsymm : ∀ i j : Fin (2 * n + 1), Mn i j = -(Mn j i)) (hMn1 : ∀ i j : Fin (2 * n + 1), (1 ≤ (i.1 : ℤ) - j.1 ∧ (i.1 : ℤ) - j.1 ≤ n) → Mn i j = 1) (hMnn1 : ∀ i j : Fin (2 * n + 1), (i.1 : ℤ) - j.1 > n → Mn i j = -1) : Mn.rank = putnam_1988_b5_solution n := sorry

For positive integers $n$, let $M_n$ be the $2n+1$ by $2n+1$ skew-symmetric matrix for which each entry in the first $n$ subdiagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Find, with proof, the rank of $M_n$. (According to one definition, the rank of a matrix is the largest $k$ such that there is a $k \times k$ submatrix with nonzero determinant.) One may note that \begin{align*} M_1&=\begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{pmatrix} \\ M_2&=\begin{pmatrix} 0 & -1 & -1 & 1 & 1 \\ 1 & 0 & -1 & -1 & 1 \\ 1 & 1 & 0 & -1 & -1 \\ -1 & 1 & 1 & 0 & -1 \\ -1 & -1 & 1 & 1 & 0 \end{pmatrix}. \end{align*}

Show that the rank of $M_n$ equals $2n$.

['linear_algebra']

theory putnam_1988_b5 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product" "HOL-Analysis.Cartesian_Space" begin definition putnam_1988_b5_solution :: "nat \<Rightarrow> nat" where "putnam_1988_b5_solution \<equiv> undefined" (* \<lambda> n :: nat. 2 * n *) theorem putnam_1988_b5: fixes n :: nat and Mn :: "real^'m^'m" and idx :: "nat \<Rightarrow> 'm" assumes npos: "n > 0" and ncard: "CARD('m) = 2 * n + 1" and idxbij: "\<forall> m :: 'm. \<exists>! i \<in> {1 .. 2 * n + 1}. idx i = m" and Mnskewsymm: "\<forall> i \<in> {1 .. 2 * n + 1}. \<forall> j \<in> {1 .. 2 * n + 1}. Mn$(idx i)$(idx j) = -(Mn$(idx j)$(idx i))" and hMn1: "\<forall> i \<in> {1 .. 2 * n + 1}. \<forall> j \<in> {1 .. 2 * n + 1}. (1 \<le> i - j \<and> i - j \<le> n) \<longrightarrow> Mn$(idx i)$(idx j) = 1" and hMnn1: "\<forall> i \<in> {1 .. 2 * n + 1}. \<forall> j \<in> {1 .. 2 * n + 1}. i - j > n \<longrightarrow> Mn$(idx i)$(idx j) = -1" shows "rank Mn = putnam_1988_b5_solution n" sorry end

putnam_1988_b6

theorem putnam_1988_b6 (trinums : Set ℤ) (htrinums : trinums = {t : ℤ | ∃ n : ℤ, t ≥ 0 ∧ t = (n * (n + 1)) / 2}) : {(a, b) : ℤ × ℤ | ∀ t > 0, (a * t + b) ∈ trinums ↔ t ∈ trinums}.encard = ⊤ := sorry

Prove that there exist an infinite number of ordered pairs $(a,b)$ of integers such that for every positive integer $t$, the number $at+b$ is a triangular number if and only if $t$ is a triangular number. (The triangular numbers are the $t_n=n(n+1)/2$ with $n$ in $\{0,1,2,\dots\}$.)

['number_theory', 'algebra']

Section putnam_1988_b6. Require Import Ensembles Finite_sets. Theorem putnam_1988_b6: let triangular (n: nat) := exists (m: nat), n = Nat.div (m * (m + 1)) 2 in exists (E: Ensemble (nat*nat)), forall (a b: nat), (E (a, b) <-> exists (m: nat), triangular m <-> triangular (Nat.mul a m + b) ) -> ~ exists (n: nat), cardinal (nat*nat) E n. Proof. Admitted. End putnam_1988_b6.

theory putnam_1988_b6 imports Complex_Main begin theorem putnam_1988_b6: fixes trinums :: "int set" defines "trinums \<equiv> {t :: int. \<exists> n :: nat. t = (n * (n + 1)) / 2}" shows "infinite {(a :: int, b :: int). \<forall> t > 0. (a * t + b) \<in> trinums \<longleftrightarrow> t \<in> trinums}" sorry end

putnam_2022_a1

abbrev putnam_2022_a1_solution : Set (ℝ × ℝ) := sorry -- {(a, b) | (a = 0 ∧ b = 0) ∨ (|a| ≥ 1) ∨ (0 < |a| ∧ |a| < 1 ∧ (b < (Real.log (1 - (1 - Real.sqrt (1 - a^2))/a))^2 - |a| * (1 - Real.sqrt (1 - a^2))/a ∨ b > (Real.log (1 - (1 + Real.sqrt (1 - a^2))/a))^2 - |a| * (1 + Real.sqrt (1 - a^2))/a))} theorem putnam_2022_a1 : {(a, b) | ∃! x : ℝ, a * x + b = Real.log (1 + x^2)} = putnam_2022_a1_solution := sorry

Determine all ordered pairs of real numbers $(a,b)$ such that the line $y = ax+b$ intersects the curve $y = \ln(1+x^2)$ in exactly one point.

Show that the solution is the set of ordered pairs $(a,b)$ which satisfy at least one of (1) $a = b = 0$, (2) $|a| \geq 1$, and (3) $0 < |a| < 1$ and $b < \log(1 - r_{-})^2 - |a|r_{-}$ or $b > \log(1 + r_{+})^2 + |a|r_{+}$ where $r_{\pm} = \frac{1 \pm \sqrt{1 - a^2}}{a}$.

['algebra']

Section putnam_2022_a1. Require Import Reals Coquelicot.Coquelicot. Definition putnam_2022_a1_solution (a b: R) := (a = 0 /\ b = 0) \/ (Rabs a >= 1) \/ (Rabs a < 1 /\ (b < ln (1 - (1 - sqrt (1 - a ^ 2)) / a) ^ 2 - Rabs a * (1 - (1 - sqrt (1 - a ^ 2)) / a) \/ b > ln (1 - (1 + sqrt (1 - a ^ 2) / a) ^ 2 - Rabs a * (1 + sqrt (1 - a ^ 2) / a)))). Theorem putnam_2022_a1 : forall a b : R, (exists! (x: R), a * x + b = ln (1 + x ^ 2) / ln 10) <-> putnam_2022_a1_solution a b. Proof. Admitted. End putnam_2022_a1.

theory putnam_2022_a1 imports Complex_Main begin definition putnam_2022_a1_solution :: "(real \<times> real) set" where "putnam_2022_a1_solution \<equiv> undefined" (* {(a, b). (a = 0 \<and> b = 0) \<or> ((abs a) \<ge> 1) \<or> (0 < (abs a) \<and> (abs a) < 1 \<and> (b < (ln (1 - (1 - sqrt (1 - a^2))/a))^2 - (abs a) * (1 - sqrt (1 - a^2))/a \<or> b > (ln (1 - (1 + sqrt (1 - a^2))/a))^2 - (abs a) * (1 + sqrt (1 - a^2))/a))} *) theorem putnam_2022_a1: shows "{(a, b). \<exists>! x :: real. a * x + b = ln (1 + x^2)} = putnam_2022_a1_solution" sorry end

putnam_2022_a2

abbrev putnam_2022_a2_solution : ℕ → ℕ := sorry -- fun n => 2*n - 2 theorem putnam_2022_a2 (n : ℕ) (hn : n ≥ 2) (S : Set ℝ[X] := {P : ℝ[X] | natDegree P = n}) (negs : ℝ[X] → ℕ := fun P : ℝ[X] => ∑ i in Finset.range (P.natDegree + 1), if P.coeff i < 0 then 1 else 0) : sSup {negs (P^2) | P ∈ S} = putnam_2022_a2_solution n := sorry

Let $n$ be an integer with $n \geq 2$. Over all real polynomials $p(x)$ of degree $n$, what is the largest possible number of negative coefficients of $p(x)^2$?

Show that the solution is $2n - 2$.

['algebra']

Section putnam_2022_a2. Require Import Basics Nat Reals Coquelicot.Coquelicot. Definition putnam_2022_a2_solution : nat -> nat := fun n => sub (mul 2 n) 2. Theorem putnam_2022_a2 (n : nat) (hn : ge n 2) (num_neg_coeff : nat -> (nat -> R) -> nat := fun n coeff => Z.to_nat (floor (sum_n (fun i => if Rlt_dec (coeff i) 0 then 1 else 0) (n + 1)))) : forall (P : R -> R) (coeff1 coeff2: nat -> R) (n: nat), (((coeff1 n <> 0 /\ P = (fun x => sum_n (fun i => coeff1 i * x ^ i) (n + 1))) /\ compose P P = (fun x => sum_n (fun i => coeff2 i * x ^ i) (n + 1))) -> ge (num_neg_coeff n coeff2) (putnam_2022_a2_solution n)) /\ (exists (P : R -> R) (coeff1 coeff2: nat -> R) (n: nat), ((coeff1 n <> 0 /\ P = (fun x => sum_n (fun i => coeff1 i * x ^ i) (n + 1))) /\ compose P P = (fun x => sum_n (fun i => coeff2 i * x ^ i) (n + 1))) -> num_neg_coeff n coeff2 = putnam_2022_a2_solution n). Proof. Admitted. End putnam_2022_a2.

theory putnam_2022_a2 imports Complex_Main "HOL-Computational_Algebra.Polynomial" begin definition putnam_2022_a2_solution :: "nat \<Rightarrow> nat" where "putnam_2022_a2_solution \<equiv> undefined" (* \<lambda> n. 2*n - 2 *) theorem putnam_2022_a2: fixes n :: "nat" and S :: "(real poly) set" and negs :: "(real poly) \<Rightarrow> nat" defines "S \<equiv> {P :: real poly. degree P = 2*n}" and "negs \<equiv> \<lambda> P :: real poly. \<Sum> i=0..(degree P). (if coeff P i < 0 then 1 else 0)" assumes hn : "n \<ge> 2" shows "(GREATEST m :: nat. \<exists> P \<in> S. negs (P^2) = m) = putnam_2022_a2_solution n" sorry end

putnam_2022_a3

theorem putnam_2022_a3 (p : ℕ) (hp : Nat.Prime p ∧ p > 5) (f : ℕ := {a : ℕ → (ZMod p) | ∀ n : ℕ, a n ≠ 0 ∧ a n * a (n + 2) = 1 + a (n + 1)}.ncard) : f ≡ 0 [MOD 5] ∨ f ≡ 2 [MOD 5] := sorry

Let $p$ be a prime number greater than 5. Let $f(p)$ denote the number of infinite sequences $a_1, a_2, a_3, \dots$ such that $a_n \in \{1, 2, \dots, p-1\}$ and $a_n a_{n+2} \equiv 1 + a_{n+1} \pmod{p}$ for all $n \geq 1$. Prove that $f(p)$ is congruent to 0 or 2 $\pmod{5}$.

['number_theory']

theory putnam_2022_a3 imports Complex_Main "HOL-Number_Theory.Cong" begin theorem putnam_2022_a3: fixes p :: "nat" and f :: "nat" defines "f \<equiv> card {a :: nat \<Rightarrow> int. \<forall> n :: nat. a n \<in> {1..(p-1)} \<and> [a n * a (n + 2) = 1 + a (n + 1)] (mod p)}" assumes hp : "prime p \<and> p > 5" shows "[f = 0] (mod 5) \<or> [f = 2] (mod 5)" sorry end

putnam_2022_a6

abbrev putnam_2022_a6_solution : ℕ → ℕ := sorry -- (fun n : ℕ => n) theorem putnam_2022_a6 (n : ℕ) (xlt : (ℕ → ℝ) → Prop) (mxsum : ℕ → (ℕ → ℝ) → Prop) (mexx : ℕ → Prop) (npos : n > 0) (hxlt : ∀ x : ℕ → ℝ, xlt x = ((-1 < x 1) ∧ (∀ i : Set.Icc 1 (2 * n - 1), x i < x (i + 1)) ∧ (x (2 * n) < 1))) (hmxsum : ∀ m : ℕ, ∀ x : ℕ → ℝ, mxsum m x = ∀ k ∈ Set.Icc 1 m, (∑ i : Fin n, ((x (2 * (i.1 + 1))) ^ (2 * k - 1) - (x (2 * (i.1 + 1) - 1)) ^ (2 * k - 1))) = 1) (hmexx : ∀ m : ℕ, mexx m = ∃ x : ℕ → ℝ, xlt x ∧ mxsum m x) : mexx (putnam_2022_a6_solution n) ∧ (∀ m : ℕ, mexx m → m ≤ putnam_2022_a6_solution n) := sorry

Let $n$ be a positive integer. Determine, in terms of $n$, the largest integer $m$ with the following property: There exist real numbers $x_1,\dots,x_{2n}$ with $-1<x_1<x_2<\cdots<x_{2n}<1$ such that the sum of the lengths of the $n$ intervals $[x_1^{2k-1},x_2^{2k-1}],[x_3^{2k-1},x_4^{2k-1}],\dots,[x_{2n-1}^{2k-1},x_{2n}^{2k-1}]$ is equal to $1$ for all integers $k$ with $1 \leq k \leq m$.

Show that the largest such $m$ is $n$.

['algebra']

Section putnam_2022_a6. Require Import Nat Reals Coquelicot.Hierarchy. From mathcomp Require Import div fintype seq ssralg ssrbool ssrnat ssrnum . Definition putnam_2022_a6_solution := fun n:nat => n. Theorem putnam_2022_a6: forall (N M n: nat), n = mul 2 N -> exists (s: 'I_n -> R) (i i0 : 'I_n), s i < s (ordS i) /\ s (nth i0 (enum 'I_n) 0) > -1 /\ s (nth i0 (enum 'I_n) (n-1)) < 1 -> forall (k: nat), and (le 1 k) (le k M) -> let valid (M : nat) : Prop := sum_n (fun i => (((s (nth i0 (enum 'I_n) (i+1))))^(2*k-1) - ((s (nth i0 (enum 'I_n) i)))^(2*k-1))) (n-1) = 1 in valid M <-> le M (putnam_2022_a6_solution n) /\ valid (putnam_2022_a6_solution n). Proof. Admitted. End putnam_2022_a6.

theory putnam_2022_a6 imports Complex_Main begin (* uses (nat \<Rightarrow> real) instead of (Fin (2*n) \<Rightarrow> real) *) definition putnam_2022_a6_solution :: "nat \<Rightarrow> nat" where "putnam_2022_a6_solution \<equiv> undefined" (* (\<lambda>n::nat. n) *) theorem putnam_2022_a6: fixes n :: nat and xlt :: "(nat \<Rightarrow> real) \<Rightarrow> bool" and mxsum :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> bool" and mexx :: "nat \<Rightarrow> bool" assumes npos: "n > 0" and hxlt: "\<forall>x::nat\<Rightarrow>real. xlt x = (-1 < x 1 \<and> (\<forall>i::nat\<in>{1..(2*n-1)}. x i < x (i + 1)) \<and> x (2*n) < 1)" and hmxsum: "\<forall>(m::nat)(x::nat\<Rightarrow>real). mxsum m x = (\<forall>k::nat\<in>{1..m}. ((\<Sum>i::nat=1..n. ((x (2*i))^(2*k-1) - (x (2*i-1))^(2*k-1))) = 1))" and hmexx: "\<forall>m::nat. mexx m = (\<exists>x::nat\<Rightarrow>real. xlt x \<and> mxsum m x)" shows "(GREATEST m::nat. mexx m) = putnam_2022_a6_solution n" sorry end

putnam_2022_b1

theorem putnam_2022_b1 (n : ℕ) (P : Polynomial ℝ) (B : Polynomial ℝ) (npos : n ≥ 1) (Pconst : P.coeff 0 = 0) (Pdegree : P.degree = n) (Pint : ∀ k : Set.Icc 1 n, P.coeff k = round (P.coeff k)) (Podd : Odd (round (P.coeff 1))) (hB : ∀ x : ℝ, Real.exp (P.eval x) = B.eval x) : ∀ k : ℕ, B.coeff k ≠ 0 := sorry

Suppose that $P(x)=a_1x+a_2x^2+\cdots+a_nx^n$ is a polynomial with integer coefficients, with $a_1$ odd. Suppose that $e^{P(x)}=b_0+b_1x+b_2x^2+\dots$ for all $x$. Prove that $b_k$ is nonzero for all $k \geq 0$.

['analysis', 'algebra']

Section putnam_2022_b1. Require Import Nat Factorial ZArith. From mathcomp Require Import fintype ssralg ssrnat ssrnum poly polydiv. Local Open Scope ring_scope. Theorem putnam_2022_b1: forall (R: numDomainType) (n : nat) (a: nat -> Z), forall (i: nat), le i n /\ (Z.odd (a 0%nat) = true) -> let p : {poly R} := \sum_(i < n) (if (0 <? (a i))%Z then ((Z.to_nat (a i))%:R *: 'X^i) else((Z.to_nat (a i))%:R *: -'X^i)) in exists (b : {poly R}), \sum_(i < n) (iter n (comp_poly p) 1) / (fact n)%:R = b -> forall (i: nat), p`_i >= 0 = true. Proof. Admitted. End putnam_2022_b1.

theory putnam_2022_b1 imports Complex_Main begin theorem putnam_2022_b1: fixes n :: nat and P :: "real poly" and B :: "real poly" assumes npos: "n \<ge> 1" and Pconst: "coeff P 0 = 0" and Pdegree: "degree P = n" and Pint: "\<forall>k::nat\<in>{1..n}. coeff P k = round (coeff P k)" and podd: "odd (round (coeff P 1))" and hB: "\<forall>x::real. exp (poly P x) = poly B x" shows "\<forall>k::nat. coeff B k \<noteq> 0" sorry end

putnam_2022_b2

abbrev putnam_2022_b2_solution : Set ℕ := sorry -- {1, 7} theorem putnam_2022_b2 (n : ℕ) (Scross : Finset (Fin 3 → ℝ) → Prop) (hScross : ∀ S : Finset (Fin 3 → ℝ), Scross S = (S = {u : Fin 3 → ℝ | ∃ v w : S, u = crossProduct v w})) : (n > 0 ∧ ∃ S : Finset (Fin 3 → ℝ), S.card = n ∧ Scross S) ↔ n ∈ putnam_2022_b2_solution := sorry

Let $\times$ represent the cross product in $\mathbb{R}^3$. For what positive integers $n$ does there exist a set $S \subset \mathbb{R}^3$ with exactly $n$ elements such that $S=\{v imes w:v,w \in S\}$?

Show that the possible values of $n$ are $1$ and $7$.

['algebra']

Section putnam_2022_b2. Require Import Ensembles Finite_sets List Reals. Require Import GeoCoq.Main.Tarski_dev.Ch16_coordinates_with_functions. Context `{T2D:Tarski_2D} `{TE:@Tarski_euclidean Tn TnEQD}. Import ListNotations. Definition vect3:= (F * F * F)%type. Definition cross_prod (v w : vect3) := let '(v1, v2, v3) := v in let '(w1, w2, w3) := w in (SubF (MulF v2 w3) (MulF v3 w2), SubF (MulF v3 w1) (MulF v1 w3), SubF (MulF v1 w2) (MulF v2 w1)). Definition putnam_2022_b2_solution := [1; 7]. Theorem putnam_2022_b2: forall (n: nat), n > 0 /\ exists (A: Ensemble vect3), forall (u: vect3), A u <-> exists (v w: vect3), u = cross_prod v w /\ cardinal vect3 A n -> In n putnam_2022_b2_solution. Proof. Admitted. End putnam_2022_b2.

theory putnam_2022_b2 imports Complex_Main begin definition putnam_2022_b2_solution :: "nat set" where "putnam_2022_b2_solution \<equiv> undefined" (* {1, 7} *) theorem putnam_2022_b2: fixes n :: nat and Scross :: "(real^3) set \<Rightarrow> bool" assumes hScross: "\<forall>S::(real^3) set. Scross S = (S = {u::real^3. (\<exists>v\<in>S. \<exists>w\<in>S. u = cross3 v w)})" shows "(n > 0 \<and> (\<exists>S::(real^3) set. finite S \<and> card S = n \<and> Scross S)) \<longleftrightarrow> n \<in> putnam_2022_b2_solution" sorry end

putnam_2022_b4

abbrev putnam_2022_b4_solution : Set ℕ := sorry -- {n : ℕ | 3 ∣ n ∧ n ≥ 9} theorem putnam_2022_b4 (n : ℕ) (ap3 : ℝ → ℝ → ℝ → Prop) (xprog : (ℕ → ℝ) → Prop) (hap3 : ∀ x0 x1 x2 : ℝ, ap3 x0 x1 x2 = ∀ o0 o1 o2 : ℝ, (o0 < o1 ∧ o1 < o2 ∧ ({o0, o1, o2} : Set ℝ) = {x0, x1, x2}) → (o1 - o0 = o2 - o1)) (hxprog : ∀ x : ℕ → ℝ, xprog x = ((∀ i j : Fin n, i.1 ≠ j.1 → x i.1 ≠ x j.1) ∧ (∀ i : Fin n, ap3 (x i.1) (x ((i.1 + 1) % n)) (x ((i.1 + 2) % n))))) : (n ≥ 4 ∧ ∃ x : ℕ → ℝ, xprog x) ↔ n ∈ putnam_2022_b4_solution := sorry

Find all integers $n$ with $n \geq 4$ for which there exists a sequence of distinct real numbers $x_1,\dots,x_n$ such that each of the sets $\{x_1,x_2,x_3\},\{x_2,x_3,x_4\},\dots,\{x_{n-2},x_{n-1},x_n\},\{x_{n-1},x_n,x_1\}$, and $\{x_n,x_1,x_2\}$ forms a $3$-term arithmetic progression when arranged in increasing order.

Show that the values of $n$ in question are the multiples of $3$ starting with $9$.

['algebra']

Section putnam_2022_b4. Require Import Reals. From mathcomp Require Import fintype seq ssrbool. Definition putnam_2022_b4_solution := fun x => (x > 9) /\ (x mod 3 = 0). Local Open Scope R. Theorem putnam_2022_b4: forall (n: nat) (s: 'I_n -> R), ge n 4 -> forall (i i0: 'I_n), (2 * (s (nth i0 (enum 'I_n) ((i-1) mod n))) = s (nth i0 (enum 'I_n) (i mod n)) + s (nth i0 (enum 'I_n) ((i+1) mod n))) \/ (2 * (s (nth i0 (enum 'I_n) (i mod n))) = s (nth i0 (enum 'I_n) ((i-1) mod n)) + s (nth i0 (enum 'I_n) ((i+1) mod n))) \/ (2 * (s (nth i0 (enum 'I_n) ((i+1) mod n))) = s (nth i0 (enum 'I_n) ((i-1) mod n)) + s (nth i0 (enum 'I_n) (i mod n))) <-> putnam_2022_b4_solution n. Proof. Admitted. End putnam_2022_b4.

theory putnam_2022_b4 imports Complex_Main begin (* uses (nat \<Rightarrow> real) instead of (Fin n \<Rightarrow> real) *) definition putnam_2022_b4_solution :: "nat set" where "putnam_2022_b4_solution \<equiv> undefined" (* {n::nat. 3 dvd n \<and> n \<ge> 9} *) theorem putnam_2022_b4: fixes n :: nat and ap3 :: "real \<Rightarrow> real \<Rightarrow> real \<Rightarrow> bool" and xprog :: "(nat \<Rightarrow> real) \<Rightarrow> bool" assumes hap3: "\<forall>x0 x1 x2::real. ap3 x0 x1 x2 = (\<forall>o0 o1 o2::real. (o0 < o1 \<and> o1 < o2 \<and> {o0, o1, o2} = {x0, x1, x2}) \<longrightarrow> (o1 - o0 = o2 - o1))" and hxprog: "\<forall>x::nat\<Rightarrow>real. xprog x = ((\<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. i \<noteq> j \<longrightarrow> x i \<noteq> x j) \<and> (\<forall>i::nat\<in>{0..(n-1)}. ap3 (x i) (x ((i + 1) mod n)) (x ((i + 2) mod n))))" shows "(n \<ge> 4 \<and> (\<exists>x::nat\<Rightarrow>real. xprog x)) \<longleftrightarrow> n \<in> putnam_2022_b4_solution" sorry end

putnam_2022_b6

abbrev putnam_2022_b6_solution : Set (Set.Ioi (0 : ℝ) → Set.Ioi (0 : ℝ)) := sorry -- {f : Set.Ioi (0 : ℝ) → Set.Ioi (0 : ℝ) | ∃ c : ℝ, c ≥ 0 ∧ ∀ x : Set.Ioi (0 : ℝ), f x = 1 / (1 + c * x)} theorem putnam_2022_b6 (f : Set.Ioi (0 : ℝ) → Set.Ioi (0 : ℝ)) (eq : Prop) (heq : eq = ∃ fr : ℝ → ℝ, (∀ x : Set.Ioi (0 : ℝ), fr x = f x) ∧ (∀ x y : Set.Ioi (0 : ℝ), fr (x * fr y) + fr (y * fr x) = 1 + fr (x + y))) : (Continuous f ∧ eq) ↔ f ∈ putnam_2022_b6_solution := sorry

Find all continuous functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that $f(xf(y))+f(yf(x))=1+f(x+y)$ for all $x,y>0$.

Show that the only such functions are the functions $f(x)=\frac{1}{1+cx}$ for some $c \geq 0$.

['analysis']

Section putnam_2022_b6. Require Import Reals. Local Open Scope R. Definition putnam_2022_b6_solution := fun (f : R -> R) => exists (c : R), c >= 0 /\ forall (x : R), x >= 0 /\ f x >= 0 -> f x = 1 / (1 + c * x). Theorem putnam_2022_b6: forall (f: R -> R) (x y: R), x > 0 /\ y > 0 /\ f x > 0 /\ f y > 0 -> f (x * f y) + f (y * f x) = 1 + f (x + y) <-> putnam_2022_b6_solution f. Proof. Admitted. End putnam_2022_b6.

theory putnam_2022_b6 imports Complex_Main begin (* uses (real \<Rightarrow> real) instead of ({0<..} \<Rightarrow> {0<..}) *) definition putnam_2022_b6_solution :: "(real \<Rightarrow> real) set" where "putnam_2022_b6_solution \<equiv> undefined" (* {f::real\<Rightarrow>real. (\<exists>c::real. c \<ge> 0 \<and> (\<forall>x::real>0. f x = 1/(1+c*x)))} *) theorem putnam_2022_b6: fixes f :: "real \<Rightarrow> real" and eq :: bool assumes heq: "eq = (\<forall>x::real>0. \<forall>y::real>0. f (x*f y) + f (y*f x) = 1 + f (x+y))" shows "(continuous_on {0<..} f \<and> eq) \<longleftrightarrow> f \<in> putnam_2022_b6_solution" sorry end

putnam_1973_a3

theorem putnam_1973_a3 (b : ℤ → ℝ := fun n => sInf {k + n/k | k > 0}) : ∀ n : ℤ, n > 0 → floor (b n) = floor (Real.sqrt (4 * n + 1)) := sorry

Let $n$ be a fixed positive integer and let $b(n)$ be the minimum value of $k + \frac{n}{k}$ as $k$ is allowed to range through all positive integers. Prove that $b(n)$ and $\sqrt{4n + 1}$ have the same integer part.

['number_theory']

theory putnam_1973_a3 imports Complex_Main begin theorem putnam_1973_a3: fixes b :: "nat \<Rightarrow> real" defines "b \<equiv> (\<lambda>n::nat. (LEAST knk::real. (\<exists>k::nat>0. knk = k + n/k)))" shows "\<forall>n::nat. (n > 0 \<longrightarrow> \<lfloor>b n\<rfloor> = \<lfloor>sqrt (4*n + 1)\<rfloor>)" sorry end

putnam_1973_a4

abbrev putnam_1973_a4_solution : ℕ := sorry -- 3 theorem putnam_1973_a4 (f : ℝ → ℝ := fun x => 2^x - 1 - x^2) : putnam_1973_a4_solution = {x : ℝ | f x = 0}.ncard := sorry

How many zeros does the function $f(x) = 2^x - 1 - x^2$ have on the real line?

Show that the solution is 3.

['analysis']

theory putnam_1973_a4 imports Complex_Main begin definition putnam_1973_a4_solution :: nat where "putnam_1973_a4_solution \<equiv> undefined" (* 3 *) theorem putnam_1973_a4: fixes f :: "real \<Rightarrow> real" defines "f \<equiv> (\<lambda>x::real. 2 powr x - 1 - x^2)" shows "putnam_1973_a4_solution = card {x::real. (f x = 0)}" sorry end

putnam_1973_a6

theorem putnam_1973_a6 (h_nint : ℕ → (Fin 7 → (ℝ × ℝ)) → ℕ := fun n lines => {p : ℝ × ℝ | ∃! S : Set (Fin 7), S.ncard = n ∧ p ∈ ⋂ i ∈ S, {pts | pts.2 = (lines i).1 * pts.1 + (lines i).2}}.ncard) : ¬ ∃ lines : Fin 7 → (ℝ × ℝ), (∀ i j : Fin 7, i ≠ j → lines i ≠ lines j) ∧ h_nint 3 lines ≥ 6 ∧ h_nint 2 lines ≥ 4 := sorry

Prove that it is impossible for seven distinct straight lines to be situated in the Euclidean plane so as to have at least six points where exactly three of these lines intersect and at least four points where exactly two of these lines interest.

['geometry']

theory putnam_1973_a6 imports Complex_Main begin (* Note: Uses the slope/intercept formulation of line in the plane *) (* uses (nat \<Rightarrow> real) instead of (Fin 7 \<Rightarrow> real) *) theorem putnam_1973_a6: fixes h_nint :: "nat \<Rightarrow> (nat \<Rightarrow> (real \<times> real)) \<Rightarrow> nat" defines "h_nint \<equiv> (\<lambda>(n::nat)(lines::nat\<Rightarrow>(real\<times>real)). card {p::real\<times>real. (\<exists>!S::nat set. S \<subseteq> {0..6} \<and> card S = n \<and> p \<in> (\<Inter>i\<in>S. {pts::real\<times>real. (snd pts = fst (lines i) * fst pts + snd (lines i))}))})" shows "\<not>(\<exists>lines::nat\<Rightarrow>(real\<times>real). (\<forall>i::nat\<in>{0..6}. \<forall>j::nat\<in>{0..6}. (i \<noteq> j \<longrightarrow> lines i \<noteq> lines j)) \<and> h_nint 3 lines \<ge> 6 \<and> h_nint 2 lines \<ge> 4)" sorry end

putnam_1973_b1

theorem putnam_1973_b1 (n : ℕ) (a : Finset.Icc 1 (2 * n + 1) → ℤ) (h_remove : ∀ S : Finset (Finset.Icc 1 (2 * n + 1)), S.card = 2*n → ∃ T, T ⊆ S ∧ T.card = n ∧ ∑ i in T, a i = ∑ i in (S \ T), a i) : ∀ i j : Finset.Icc 1 (2 * n + 1), a i = a j := sorry

Let $a_1, \dots, a_{2n + 1}$ be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of $n$ integers with equal sums. Prove $a_1 = a_2 = \dots = a_{2n+1}$.

['algebra']

theory putnam_1973_b1 imports Complex_Main begin (* uses (nat \<Rightarrow> int) instead of ({1..(2*n+1)} \<Rightarrow> int) *) theorem putnam_1973_b1: fixes n :: nat and a :: "nat \<Rightarrow> int" assumes h_remove: "\<forall>S::nat set. ((S \<subseteq> {1..(2*n+1)} \<and> card S = 2*n) \<longrightarrow> (\<exists>T::nat set. T \<subseteq> S \<and> card T = n \<and> (\<Sum>i\<in>T. a i) = (\<Sum>i\<in>(S-T). a i)))" shows "\<forall>i::nat\<in>{1..(2*n+1)}. \<forall>j::nat\<in>{1..(2*n+1)}. a i = a j" sorry end

putnam_1973_b2

theorem putnam_1973_b2 (z : ℂ) (hzrat : ∃ q1 q2 : ℚ, z.re = q1 ∧ z.im = q2) (hznorm : ‖z‖ = 1) : ∀ n : ℤ, ∃ q : ℚ, ‖z^(2*n) - 1‖ = q := sorry

Let $z = x+iy$ be a complex number with $x$ and $y$ rational and with $\| z \| = 1$. Show thaat the number $\| z^{2n} - 1 \|$ is rational for every integer $n$.

['number_theory']

theory putnam_1973_b2 imports Complex_Main begin theorem putnam_1973_b2: fixes z :: complex assumes hzrat: "\<exists>q1 q2::rat. Re z = q1 \<and> Im z = q2" and hznorm: "norm z = 1" shows "\<forall>n::int. (\<exists>q::rat. norm (z powi (2*n) - 1) = q)" sorry end

putnam_1973_b3

theorem putnam_1973_b3 (p : ℕ) (pgt1 : p > 1) (hprime : ∀ x ∈ Set.Ico 0 p, Nat.Prime (x^2 - x + p)) : ∃! triple : ℤ × ℤ × ℤ, let (a,b,c) := triple; b^2 - 4*a*c = 1 - 4*p ∧ 0 < a ∧ a ≤ c ∧ -a ≤ b ∧ b < a := sorry

Let $p > 1$ be an integer with the property that $x^2 - x + p$ is prime for all $x$ in the range $0 < x < p$. Show there exists exactly one triple of integers $a,b,c$ satisfying $b^2 - 4ac = 1 - 4p$, $0 < a \leq c$, and $-a \leq b < a$.

['number_theory', 'algebra']

theory putnam_1973_b3 imports Complex_Main "HOL-Computational_Algebra.Primes" begin theorem putnam_1973_b3: fixes p :: nat assumes pgt1: "p > 1" and hprime: "\<forall>x::nat\<in>{0..<p}. prime (x^2 - x + p)" shows "\<exists>!triple::int\<times>int\<times>int. let (a,b,c) = triple in (b^2 - 4*a*c = 1 - 4*p \<and> 0 < a \<and> a \<le> c \<and> -a \<le> b \<and> b < a)" sorry end

putnam_1973_b4

abbrev putnam_1973_b4_solution : ℝ → ℝ := sorry -- (fun x => x) theorem putnam_1973_b4 (f : ℝ → ℝ) (hprop : (ℝ → ℝ) → Prop := fun g => ContDiff ℝ 1 g ∧ (∀ x : ℝ, 0 < deriv g x ∧ deriv g x ≤ 1) ∧ g 0 = 0) (hf : hprop f) : (∫ x in Icc 0 1, f x)^2 ≥ ∫ x in Icc 0 1, (f x)^3 ∧ (hprop putnam_1973_b4_solution ∧ (∫ x in Icc 0 1, putnam_1973_b4_solution x)^2 = ∫ x in Icc 0 1, (putnam_1973_b4_solution x)^3) := sorry

Suppose $f$ is a function on $[0,1]$ with continuous derivative satisfying $0 < f'(x) \leq 1$ and $f 0 = 0$. Prove that $\left[\int_0^1 f(x) dx\right]]^2 \geq \int_0^1 (f(x))^3 dx$, and find an example where equality holds.

Show that one such example where equality holds is the identity function.

['analysis']

theory putnam_1973_b4 imports Complex_Main "HOL-Analysis.Interval_Integral" begin (* Note: Boosted domain to real, which is fine because you can extend any such function f from [0,1] to real satisfying the same properties. *) (* Note: There may be multiple correct answers. *) definition putnam_1973_b4_solution :: "real \<Rightarrow> real" where "putnam_1973_b4_solution \<equiv> undefined" (* (\<lambda>x::real. x) *) theorem putnam_1973_b4: fixes f :: "real \<Rightarrow> real" and hprop :: "(real \<Rightarrow> real) \<Rightarrow> bool" defines "hprop \<equiv> (\<lambda>g::real\<Rightarrow>real. g C1_differentiable_on UNIV \<and> (\<forall>x::real. 0 < deriv g x \<and> deriv g x \<le> 1) \<and> g 0 = 0)" assumes hf: "hprop f" shows "((interval_lebesgue_integral lebesgue 0 1 f)^2 \<ge> interval_lebesgue_integral lebesgue 0 1 (\<lambda>x::real. (f x)^3)) \<and> (hprop putnam_1973_b4_solution \<and> ((interval_lebesgue_integral lebesgue 0 1 putnam_1973_b4_solution)^2 = interval_lebesgue_integral lebesgue 0 1 (\<lambda>x::real. (putnam_1973_b4_solution x)^3)))" sorry end