theory "Ireland-Rosen"
  imports "HOL-Number_Theory.Number_Theory"
          "Dirichlet_Series.Moebius_Mu"
          "Gaussian_Integers.Gaussian_Integers_Everything"

begin

(*
problem_number:1_27
natural language statement:
For all odd $n$ show that $8 \mid n^{2}-1$.
lean statement:
theorem exercise_1_27 {n : \<nat>} (hn : odd n) : 8 dvd (n^2 - 1) :=

codex statement:
theorem div_8_of_odd_n:
  fixes n::nat
  assumes "odd n"
  shows "8 dvd n^2 - 1"
Our comment on the codex statement: perfect
 *)
theorem exercise_1_27:
  fixes n::nat
  assumes "odd n"
  shows "8 dvd n^2 - 1"
  oops

(*
problem_number:1_30
natural language statement:
Prove that $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is not an integer.
lean statement:
theorem exercise_1_30 {n : \<nat>} :
  \<not> \<exists> a : \<int>, \<Sum> (i : fin n), (1 : \<rat>) / (n+2) = a :=

codex statement:
theorem sum_frac_not_integer:
  fixes n::nat
  shows "\<Sum>i<n. 1 / real i \<noteq> int (\<Sum>i<n. 1 / real i)"
Our comment on the codex statement: The right side was nonsense
 *)
theorem exercise_1_30:
  fixes n::nat
  assumes "n \<ge> 2"
  shows "(\<Sum>i=2..n. 1 / real i) \<notin> \<int>"
  oops

(*
problem_number:1_31
natural language statement:
Show that 2 is divisible by $(1+i)^{2}$ in $\mathbb{Z}[i]$.
lean statement:
theorem exercise_1_31  : (\<langle>1, 1\<rangle> : gaussian_int) ^ 2 dvd 2 :=

codex statement:
theorem divisible_of_power_of_one_plus_i:
  fixes z::complex
  assumes "z \<in> {x + y * \<i> |x y. x \<in> UNIV \<and> y \<in> UNIV}"
  shows "\<exists>a. 2 = (1 + \<i>)^2 * a"
Our comment on the codex statement: Confused by the notation it assumed the complex numbers
 *)
theorem exercise_1_31: "(1 + \<i>\<^sub>\<int>)^2 dvd 2"
  by (simp add: gauss_int_2_eq)

(*
problem_number:2_4
natural language statement:
If $a$ is a nonzero integer, then for $n>m$ show that $\left(a^{2^{n}}+1, a^{2^{m}}+1\right)=1$ or 2 depending on whether $a$ is odd or even.
lean statement:
theorem exercise_2_4 {a : \<int>} (ha : a \<noteq> 0)
  (f_a := \<lambda> n m : \<nat>, int.gcd (a^(2^n) + 1) (a^(2^m)+1)) {n m : \<nat>}
  (hnm : n > m) :
  (odd a \<rightarrow> f_a n m = 1) \<and> (even a \<rightarrow> f_a n m = 2) :=

codex statement:
theorem gcd_of_power_of_two_plus_one_eq_one_or_two:
  fixes a::int
  assumes "a \<noteq> 0" "n > m"
  shows "gcd (a^(2^n) + 1) (a^(2^m) + 1) = 1 \<or> gcd (a^(2^n) + 1) (a^(2^m) + 1) = 2"
Our comment on the codex statement: almost correct but overlooked the dependence on the parity of a
(which is the opposite of that suggested in the text)
 *)
theorem exercise_2_4:
  fixes a::int
  assumes "a \<noteq> 0" "n > m"
  shows "gcd (a^(2^n) + 1) (a^(2^m) + 1) = 1 \<or> gcd (a^(2^n) + 1) (a^(2^m) + 1) = (if odd a then 2 else 1)"
  oops

(*
problem_number:2_21
natural language statement:
Define $\wedge(n)=\log p$ if $n$ is a power of $p$ and zero otherwise. Prove that $\sum_{A \mid n} \mu(n / d) \log d$ $=\wedge(n)$.
lean statement:
theorem exercise_2_21 {l : \<nat> \<rightarrow> \<real>}
  (hl : \<forall> p n : \<nat>, p.prime \<rightarrow> l (p^n) = log p )
  (hl1 : \<forall> m : \<nat>, \<not> is_prime_pow m \<rightarrow> l m = 0) :
  l = \<lambda> n, \<Sum> d : divisors n, moebius (n/d) * log d  :=

codex statement:
theorem sum_mu_log_divisors_eq_log_prime_power:
  fixes n::nat
  assumes "\<exists>p. prime p \<and> n = p^a"
  shows "(\<Sum>d dvd n. \<mu> (n div d) * log d) = log p"
Our comment on the codex statement: right in places but it didn't grasp that a function was being defined
 *)
theorem exercise_2_21:
  fixes p::nat
  assumes "prime p"
  defines "L \<equiv> \<lambda>n. if \<exists>k. n = p^k then ln p else 0"
  shows "(\<Sum>d | d dvd real n. moebius_mu (n div d) * ln d) = L n"
  oops

(*
problem_number:2_27a
natural language statement:
Show that $\sum^{\prime} 1 / n$, the sum being over square free integers, diverges.
lean statement:
theorem exercise_2_27a :
  \<not> summable (\<lambda> i : {p : \<int> // squarefree p}, (1 : \<rat>) / i) :=

codex statement:
theorem diverges_sum_of_inverse_square_free_integers:
  fixes n::nat
  shows "\<Sum>i=1..n. 1 / (real_of_nat i) = \<infinity>"
Our comment on the codex statement: Overlooked of the square-free, for starters
 *)
theorem exercise_2_27a:
  "\<not> summable (\<lambda>n. if squarefree n then 1 / real n else 0)"
  oops

(*
problem_number:3_1
natural language statement:
Show that there are infinitely many primes congruent to $-1$ modulo 6 .
lean statement:
theorem exercise_3_1 : infinite {p : primes // p \<equiv> -1 [ZMOD 6]} :=

codex statement:
theorem infinite_primes_congruent_minus_one_mod_six:
  shows "infinite {p::nat. prime p \<and> p \<equiv> -1 [MOD 6]}"
Our comment on the codex statement:  not bad but the syntax was wrong and we need the integers
 *)
theorem exercise_3_1:
  shows "infinite {p::int. prime p \<and> [p = -1] (mod 6)}"
  oops

(*
problem_number:3_4
natural language statement:
Show that the equation $3 x^{2}+2=y^{2}$ has no solution in integers.
lean statement:
theorem exercise_3_4 : \<not> \<exists> x y : \<int>, 3*x^2 + 2 = y^2 :=

codex statement:
theorem no_solution_of_3_x_square_plus_2_eq_y_square:
  fixes x y::int
  assumes "3 * x^2 + 2 = y^2"
  shows "False"
Our comment on the codex statement:  it's correct however the Lean version, with existential quantifiers, may be more natural
 *)
theorem exercise_3_4:
  fixes x y::int
  assumes "3 * x^2 + 2 = y^2"
  shows "False"
  oops

(*
problem_number:3_5
natural language statement:
Show that the equation $7 x^{3}+2=y^{3}$ has no solution in integers.
lean statement:
theorem exercise_3_5 : \<not> \<exists> x y : \<int>, 7*x^3 + 2 = y^3 :=

codex statement:
theorem no_solution_of_eq_7_x_cube_plus_2_eq_y_cube:
  assumes "\<exists>x y. 7 * x^3 + 2 = y^3"
  shows False
Our comment on the codex statement: should have been treated the same as the previous one. This time we use quantifiers.
 *)
theorem exercise_3_5:
  shows "\<nexists>x y::int. 7 * x^3 + 2 = y^3"
  oops

(*
problem_number:3_10
natural language statement:
If $n$ is not a prime, show that $(n-1) ! \equiv 0(n)$, except when $n=4$.
lean statement:
theorem exercise_3_10 {n : \<nat>} (hn0 : \<not> n.prime) (hn1 : n \<noteq> 4) :
  factorial (n-1) \<equiv> 0 [MOD n] :=

codex statement:
theorem fact_minus_one_equiv_zero_of_not_prime:
  fixes n::nat
  assumes "n>2" "\<forall>p. prime p \<longrightarrow> p dvd n \<longrightarrow> p = 2 \<or> p = n"
  shows "(fact (n - 1)) mod n = 0"
Our comment on the codex statement:  completely wrong. And the text is wrong to because n needs to be non-zero
 *)
theorem exercise_3_10:
  fixes n::nat
  assumes "\<not> prime n" "n\<noteq>4" "n\<noteq>0"
  shows "[fact (n - 1) = 0] (mod n)"
  oops

(*
problem_number:3_14
natural language statement:
Let $p$ and $q$ be distinct odd primes such that $p-1$ divides $q-1$. If $(n, p q)=1$, show that $n^{q-1} \equiv 1(p q)$.
lean statement:
theorem exercise_3_14 {p q n : \<nat>} (hp0 : p.prime \<and> p > 2)
  (hq0 : q.prime \<and> q > 2) (hpq0 : p \<noteq> q) (hpq1 : p - 1 dvd q - 1)
  (hn : n.gcd (p*q) = 1) :
  n^(q-1) \<equiv> 1 [MOD p*q] :=

codex statement:
theorem congruent_one_of_prime_diff_one_divides_prime_diff_one:
  fixes p q n::nat
  assumes "prime p" "prime q" "p \<noteq> q" "p - 1 dvd q - 1" "coprime n (p * q)"
  shows "n [^] (q - 1) \<equiv> 1 (mod p * q)"
Our comment on the codex statement:  uses group exponentiation rather than numeric exponentiation. The primes do not need to be odd!
 *)
theorem exercise_3_14:
  fixes p q n::nat
  assumes "prime p" "prime q" "p \<noteq> q" "odd p" "odd q" "(p - 1) dvd (q - 1)" "coprime n (p * q)"
  shows "[n^(q-1) = 1] (mod p * q)"
  by (metis Carmichael_divides Carmichael_prime assms(1-3) assms(6-7) coprime_common_divisor coprime_commute coprime_cong_mult_nat coprime_mult_right_iff dvd_triv_right fermat_theorem not_prime_unit primes_coprime)

(*
problem_number:3_18
natural language statement:
Let $N$ be the number of solutions to $f(x) \equiv 0(n)$ and $N_{i}$ be the number of solutions to $f(x) \equiv 0\left(p_{i}^{a_{i}}\right)$. Prove that $N=N_{1} N_{2} \cdots N_{i}$.
lean statement:

codex statement:
theorem number_of_solutions_of_congruence_eq_product_of_number_of_solutions_of_congruence_mod_prime_power:
  fixes f::"nat \<Rightarrow> nat" and n::nat
  assumes "\<forall>p. prime p \<longrightarrow> \<exists>a. p^a dvd n"
  shows "\<exists>N. \<forall>p. prime p \<longrightarrow> \<exists>a. p^a dvd n \<longrightarrow> (\<exists>Np. \<forall>x. f x \<equiv> 0 [MOD p^a] \<longrightarrow> (\<exists>!x. f x \<equiv> 0 [MOD p^a])) \<longrightarrow> (\<exists>Np. \<forall>x. f x \<equiv> 0 [MOD n] \<longrightarrow> (\<exists>!x. f x \<equiv> 0 [MOD n]))"
Our comment on the codex statement:  Just wrong
 *)
theorem exercise_3_18:
  fixes f::"nat \<Rightarrow> nat" and n::nat
  defines "N \<equiv> card {x. [f x = 0] (mod n)}"
  defines "M \<equiv> {k. \<exists>p a. prime p \<and> k = card {x. [f x = 0] (mod p ^ a)}}" (* not quite right: notation needs clarifying*)
  shows "N = \<Prod>M"
  oops

(*
problem_number:3_20
natural language statement:
Show that $x^{2} \equiv 1\left(2^{b}\right)$ has one solution if $b=1$, two solutions if $b=2$, and four solutions if $b \geq 3$.
lean statement:

codex statement:
theorem number_of_solutions_of_x_square_equiv_1_mod_2_power_b:
  fixes b::nat
  assumes "b\<ge>1"
  shows "card {x::int. x^2 \<equiv> 1 [MOD 2^b]} = 2^(b-1)"
Our comment on the codex statement:  incorrect syntax and the right hand side is scrambled. The assumption b\<ge>1 is necessary and not mentioned in the problem, well done!
 *)
theorem exercise_3_20:
  fixes b::nat
  assumes "b\<ge>1"
  shows "card {x::int. [x^2 = 1] (mod 2^b)} = (if b=1 then 1 else if b=2 then 2 else 4)"
 oops

(*
problem_number:4_4
natural language statement:
Consider a prime $p$ of the form $4 t+1$. Show that $a$ is a primitive root modulo $p$ iff - $a$ is a primitive root modulo $p$.
lean statement:
theorem exercise_4_4 {p t: \<nat>} (hp0 : p.prime) (hp1 : p = 4*t + 1)
  (a : zmod p):
  is_primitive_root a p \<longleftrightarrow> is_primitive_root (-a) p :=

codex statement:
theorem primitive_root_of_prime_of_form_4_t_1:
  fixes p::nat and a::int
  assumes "prime p" "p = 4 * t + 1" "coprime a p"
  shows "\<exists>n. a [^] n \<equiv> -1 [mod p]"
Our comment on the codex statement: wrong.
 *)
theorem exercise_4_4:
  fixes p::nat and a::nat
  assumes "prime p" "[p = 1] (mod 4)"
  shows "residue_primroot p a \<longleftrightarrow> residue_primroot p (p-a)"   (*not sure this is the right way to do -a*)
  oops

(*
problem_number:4_5
natural language statement:
Consider a prime $p$ of the form $4 t+3$. Show that $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1) / 2$.
lean statement:

codex statement:
theorem primitive_root_iff_order_half_minus_a:
  fixes p::nat and a::int
  assumes "prime p" "p \<equiv> 3 (mod 4)"
  shows "\<exists>b. order b p = (p - 1) div 2 \<longleftrightarrow> primitive_root a p"
Our comment on the codex statement:  something wrong with the quantification
 *)
theorem exercise_4_5:
  fixes p::nat and a::nat
  assumes "prime p" "[p = 3] (mod 4)"
  shows "residue_primroot p a \<longleftrightarrow> ord p (p-a) = (p-1) div 2"   (*not sure this is the right way to do -a*)
  oops

(*
problem_number:4_6
natural language statement:
If $p=2^{n}+1$ is a Fermat prime, show that 3 is a primitive root modulo $p$.
lean statement:

codex statement:
theorem primitive_root_of_fermat_prime:
  fixes p::nat
  assumes "prime p" "\<exists>n. p = 2^n + 1"
  shows "\<exists>m. \<forall>k. coprime k p \<longrightarrow> (3::nat) [^] k \<equiv> 1 [^] m [MOD p]"
Our comment on the codex statement:  The conclusion looks completely wrong
 *)
theorem exercise_4_6: (* this actually fails if p=3 *)
    fixes p::nat
  assumes "prime p" "\<exists>n>0. p = 2^n + 1"
  shows "residue_primroot p 3"
oops

(*
problem_number:4_8
natural language statement:
Let $p$ be an odd prime. Show that $a$ is a primitive root module $p$ iff $a^{(p-1) / q} \not \equiv 1(p)$ for all prime divisors $q$ of $p-1$.
lean statement:

codex statement:
theorem primitive_root_of_prime_iff_not_cong_one_of_prime_divisor:
  fixes p::nat and a::int
  assumes "prime p" "odd p"
  shows "\<forall>q. prime q \<and> q dvd p - 1 \<longrightarrow> a [^] ((p - 1) div q) \<equiv> 1 (mod p) \<longleftrightarrow> \<not> primitive_root p a"
Our comment on the codex statement:  somewhat scrambled
 *)
theorem exercise_4_8:
  fixes p::nat
  assumes "prime p" "odd p"
  shows "residue_primroot p a \<longleftrightarrow> (\<forall>q. prime q \<longrightarrow> q dvd (p-1) \<longrightarrow> \<not> [a ^ ((p-1)div q) = 1] (mod p))"
  oops

(*
problem_number:4_9
natural language statement:
Show that the product of all the primitive roots modulo $p$ is congruent to $(-1)^{\phi(p-1)}$ modulo $p$.
lean statement:

codex statement:
theorem prod_primitive_roots_eq_pow_phi_minus_one_mod_p:
  fixes p::nat
  assumes "prime p"
  shows "(\<Prod>x\<in>primitive_roots p. x) \<equiv> (-1) ^ (\<phi> p - 1) [MOD p]"
Our comment on the codex statement:  not bad.
 *)
theorem exercise_4_9:
  fixes p::nat
  assumes "prime p"
  shows "[\<Prod>{r. residue_primroot p r} = (-1) ^ (totient p - 1)] (mod p)"
  oops

(*
problem_number:4_10
natural language statement:
Show that the sum of all the primitive roots modulo $p$ is congruent to $\mu(p-1)$ modulo $p$.
lean statement:

codex statement:
theorem sum_primitive_roots_eq_mu_p_minus_one:
  fixes p::nat
  assumes "prime p"
  shows "(\<Sum>x\<in>{x. x < p \<and> coprime x p \<and> \<exists>n. x [^] n \<equiv> 1 [MOD p]}. x) \<equiv> \<mu> (p - 1) [MOD p]"
Our comment on the codex statement:  it would be better not to expand the definition of primitive root
 *)
theorem exercise_4_10:
    fixes p::nat
  assumes "prime p"
  shows "[\<Sum>{r. residue_primroot p r} = moebius_mu (p - 1)] (mod p)"
  oops

(*
problem_number:4_11
natural language statement:
Prove that $1^{k}+2^{k}+\cdots+(p-1)^{k} \equiv 0(p)$ if $p-1 \nmid k$ and $-1(p)$ if $p-1 \mid k$.
lean statement:

codex statement:
theorem sum_of_powers_cong_zero_mod_p_of_p_minus_one_not_dvd_k:
  fixes p::nat and k::int
  assumes "prime p" "p-1 \<noteq> 0" "p-1 dvd k"
  shows "(\<Sum>i<p-1. i^k) \<equiv> -1 (mod p)"
Our comment on the codex statement:  It never seems to cope with statements involving two cases
 *)
theorem exercise_4_11:
  fixes p::nat
  assumes "prime p" "k \<ge> 1"
  shows "[(\<Sum>i<p. i^k) = (if (p-1) dvd k then -1 else 0)] (mod p)"
  oops

(*
problem_number:4_22
natural language statement:
If $a$ has order 3 modulo $p$, show that $1+a$ has order 6 .
lean statement:

codex statement:
theorem order_of_one_plus_a_eq_6_of_order_a_eq_3:
  fixes a::nat and p::nat
  assumes "prime p" "a mod p \<noteq> 0" "\<forall>n. a [^] n mod p \<noteq> 1 \<longrightarrow> n \<ge> 3"
  shows "\<forall>n. (1 + a) [^] n mod p \<noteq> 1 \<longrightarrow> n \<ge> 6"
Our comment on the codex statement:  it's scrambled and doesn't know about ord
 *)
theorem exercise_4_22:
  fixes a::nat and p::nat
  assumes "prime p" "ord p a = 3"
  shows "ord p (Suc a) = 6"
  oops

(*
problem_number:4_24
natural language statement:
Show that $a x^{m}+b y^{n} \equiv c(p)$ has the same number of solutions as $a x^{m^{\prime}}+b y^{n^{\prime}} \equiv c(p)$, where $m^{\prime}=(m, p-1)$ and $n^{\prime}=(n, p-1)$.
lean statement:

codex statement:
theorem number_of_solutions_of_congruence_eq_number_of_solutions_of_congruence_of_coprime_powers:
  fixes a b c::"int" and m n m' n'::nat
  assumes "prime p" "coprime m (p - 1)" "coprime n (p - 1)" "m' = gcd m (p - 1)" "n' = gcd n (p - 1)"
  shows "card {(x, y). x < p \<and> y < p \<and> a * x ^ m + b * y ^ n \<equiv> c (mod p)} = card {(x, y). x < p \<and> y < p \<and> a * x ^ m' + b * y ^ n' \<equiv> c (mod p)}"
Our comment on the codex statement:  pretty good as regards syntax except that the co-prime assumptions make no sense
 *)
theorem exercise_4_24:
  fixes m n
  assumes "prime p"
  defines "m' \<equiv> gcd m (p-1)"
  defines "n' \<equiv> gcd n (p-1)"
  shows "card {(x, y). x < p \<and> y < p \<and> [a * x ^ m + b * y ^ n = c] (mod p)} = card {(x, y). x < p \<and> y < p \<and> [a * x ^ m' + b * y ^ n' = c] (mod p)}"
  oops

(*
problem_number:5_2
natural language statement:
Show that the number of solutions to $x^{2} \equiv a(p)$ is given by $1+(a / p)$.
lean statement:
theorem exercise_5_28 {p : \<nat>} (hp : p.prime) (hp1 : p \<equiv> 1 [MOD 4]):
  \<exists> x, x^4 \<equiv> 2 [MOD p] \<longleftrightarrow> \<exists> A B, p = A^2 + 64*B^2 :=

codex statement:
theorem number_of_solutions_of_congruence_eq_one_plus_a_div_p:
  fixes a p::nat
  assumes "prime p"
  shows "card {x. x^2 \<equiv> a [MOD p]} = 1 + (a div p)"
Our comment on the codex statement: Looks good, but doesn't p have to divide a?
 *)
theorem exercise_5_2:
  fixes a p::nat
  assumes "prime p"
  shows "card {x. x < p \<and> [x^2 = a] (mod p)} = 1 + (a div p)"
  oops

(*
problem_number:5_3
natural language statement:
Suppose that $p \nmid a$. Show that the number of solutions to $a x^{2}+b x+c \equiv 0(p)$ is given by $1+\left(\left(b^{2}-4 a c\right) / p\right)$.
lean statement:
theorem exercise_5_37 {p q : \<nat>} [fact(p.prime)] [fact(q.prime)] {a : \<int>}
  (ha : a < 0) (h0 : p \<equiv> q [ZMOD 4*a]) (h1 : \<not> ((p : \<int>) dvd a)) :
  legendre_sym p a = legendre_sym q a :=

codex statement:
theorem number_of_solutions_of_quadratic_congruence_eq_1_plus_Legendre_symbol:
  fixes a b c p::nat
  assumes "prime p" "coprime a p"
  shows "card {x. x < p \<and> (a * x^2 + b * x + c) \<equiv> 0 [MOD p]} = 1 + (Legendre_symbol (a * b^2 - 4 * a * c) p)"
Our comment on the codex statement: the problem indeed looks incorrect, but is this really a use of the Legendre symbol?
 *)
theorem exercise_5_3:
  fixes p::nat
  assumes "prime p" "coprime a p"
  shows "card {x. x < p \<and> [a * x^2 + b * x + c = 0] (mod p)} = 1 + Legendre (b^2 - 4 * a * c) p"
  oops

(*
problem_number:5_4
natural language statement:
Prove that $\sum_{a=1}^{p-1}(a / p)=0$.
lean statement:

codex statement:
theorem sum_of_p_minus_one_eq_zero:
  fixes p::nat
  assumes "prime p"
  shows "(\<Sum>a=1..p-1. a / p) = 0"
Our comment on the codex statement: here it did not detect the Legendre symbol
 *)
theorem exercise_5_4:
  fixes p::nat
  assumes "prime p"
  shows "(\<Sum>a=1..p-1. Legendre a p) = 0"
  oops

(*
problem_number:5_5
natural language statement:
Prove that $\sum_{\substack{p-1 \\ x=0}}((a x+b) / p)=0$ provided that $p \nmid a .$
lean statement:

codex statement:
theorem sum_of_arithmetic_progression_eq_zero_of_coprime:
  fixes a b p::nat
  assumes "coprime a p"
  shows "(\<Sum>x=0..p-1. (a * x + b) / p) = 0"
Our comment on the codex statement: here it did not detect the Legendre symbol
 *)
theorem exercise_5_5:
    fixes p::nat
  assumes "prime p" "coprime a p"
  shows "(\<Sum>x<p. Legendre (a * x + b) p) = 0"
oops

(*
problem_number:5_6
natural language statement:
Show that the number of solutions to $x^{2}-y^{2} \equiv a(p)$ is given by $\sum_{y=0}^{p-1}\left(1+\left(\left(y^{2}+a\right) / p\right)\right) .$
lean statement:

codex statement:
theorem sum_of_solutions_of_congruence_eq_sum_of_legendre_symbol:
  fixes p::nat and a::int
  assumes "prime p"
  shows "card {x. x^2 \<equiv> a [MOD p]} = (\<Sum>y<p. 1 + legendre_symbol (y^2 + a) p)"
Our comment on the codex statement: overlooked y and used the wrong syntax for modular equivalence
 *)
theorem exercise_5_6:
  fixes p::nat and a::int
  assumes "prime p"
  shows "card {(x,y). [x^2 - y^2 = a] (mod p)} = (\<Sum>y<p. 1 + Legendre (y^2 + a) p)"
oops

(*
problem_number:5_7
natural language statement:
By calculating directly show that the number of solutions to $x^{2}-y^{2} \equiv a(p)$ is $p-1$ if $p \nmid a$ and $2 p-1$ if $p \mid a$.
lean statement:

codex statement:
theorem number_of_solutions_of_congruence_eq_p_minus_one_if_p_not_dvd_a:
  fixes p::nat and a::int
  assumes "prime p" "p dvd a"
  shows "card {x. x^2 \<equiv> a [MOD p]} = 2*p - 1"
Our comment on the codex statement:  same as in the previous case
 *)
theorem exercise_5_7:
  fixes p::nat and a::int
  assumes "prime p"
  shows "card {(x,y). [x^2 - y^2 = a] (mod p)} = (if p dvd a then 2*p - 1 else p-1)"
oops

(*
problem_number:5_13
natural language statement:
Show that any prime divisor of $x^{4}-x^{2}+1$ is congruent to 1 modulo 12 .
lean statement:
theorem exercise_5_13 {p x: \<int>} (hp : prime p)
  (hpx : p dvd (x^4 - x^2 + 1)) : p \<equiv> 1 [ZMOD 12] :=

codex statement:
theorem prime_divisor_of_x_4_minus_x_2_plus_1_cong_1_mod_12:
  fixes p::nat
  assumes "prime p" "p dvd (x^4 - x^2 + 1)"
  shows "p \<equiv> 1 (mod 12)"
Our comment on the codex statement: wrong syntax for modular equivalence
 *)
theorem exercise_5_13:
  fixes p::nat
  assumes "prime p" "p dvd x^4 - x^2 + 1"
  shows "[p=1] (mod 12)"
  oops

(*
problem_number:5_27
natural language statement:
Suppose that $f$ is such that $b \equiv a f(p)$. Show that $f^{2} \equiv-1(p)$ and that $2^{(p-1) / 4} \equiv$ $f^{a b / 2}(p)$
lean statement:

codex statement:
theorem square_eq_minus_one_of_equiv_a_f_p:
  fixes f::"int \<Rightarrow> int" and a b p::int
  assumes "\<forall>x. f (f x) = -1" "b \<equiv> a * f p [MOD p]"
  shows "f p \<equiv> a * b / 2 [MOD p]"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_5_27:   (* THIS PROBLEM IS NOT CORRECT*)
  fixes p::nat
  assumes "prime p" "[b = a*f] (mod p)"
  shows "[f^2 = -1] (mod p)"
 oops

(*
problem_number:5_28
natural language statement:
Show that $x^{4} \equiv 2(p)$ has a solution for $p \equiv 1(4)$ iff $p$ is of the form $A^{2}+64 B^{2}$.
lean statement:
theorem exercise_5_28 {p : \<nat>} (hp : p.prime) (hp1 : p \<equiv> 1 [MOD 4]):
  \<exists> x, x^4 \<equiv> 2 [MOD p] \<longleftrightarrow> \<exists> A B, p = A^2 + 64*B^2 :=

codex statement:
theorem exists_solution_of_x_4_eq_2_of_p_1_4:
  fixes p::nat
  assumes "prime p" "p \<equiv> 1 (mod 4)"
  shows "\<exists>x. x^4 \<equiv> 2 (mod p)"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_5_28:
  fixes p::nat
  assumes "prime p" "[p=1] (mod 4)"
  shows "(\<exists>x. [x^4 = 2] (mod p)) \<longleftrightarrow> (\<exists>A B. p = A^2 + 64*B^2)"
  oops

(*
problem_number:5_37
natural language statement:
Show that if $a$ is negative then $p \equiv q(4 a), p \times a$ implies $(a / p)=(a / q)$.
lean statement:
theorem exercise_5_37 {p q : \<nat>} [fact(p.prime)] [fact(q.prime)] {a : \<int>}
  (ha : a < 0) (h0 : p \<equiv> q [ZMOD 4*a]) (h1 : \<not> ((p : \<int>) dvd a)) :
  legendre_sym p a = legendre_sym q a :=

codex statement:
theorem eq_of_div_of_eq_div_of_eq_of_mult_eq_zero:
  fixes a p q::int
  assumes "a < 0" "p \<equiv> q (4 * a)" "p * a = 0"
  shows "a / p = a / q"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_5_37:
  fixes a p q::int
  assumes "a < 0" "[p=q] (mod 4*a)" "coprime p a"
  shows "Legendre a p = Legendre a q"
  oops

(*
problem_number:6_18
natural language statement:
Show that there exist algebraic numbers of arbitrarily high degree.
lean statement:

codex statement:
theorem exists_algebraic_of_degree_succ:
  fixes n::nat
  shows "\<exists>x::'a::algebra_1. algebraic x \<and> degree x = n"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_6_18: undefined oops

(*
problem_number:7_6
natural language statement:
Let $K \supset F$ be finite fields with $[K: F]=3$. Show that if $\alpha \in F$ is not a square in $F$, it is not a square in $K$.
lean statement:

codex statement:
theorem not_square_of_not_square_of_degree_three:
  fixes F::"'a::field" and K::"'b::field"
  assumes "finite_field F" "finite_field K" "F \<subseteq> K" "[K:F] = 3" "\<forall>x\<in>F. x^2 \<noteq> \<alpha>"
  shows "\<forall>x\<in>K. x^2 \<noteq> \<alpha>"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_7_6: undefined oops

(*
problem_number:7_24
natural language statement:
Suppose that $f(x) \in \mathbb{Z} / p \mathbb{Z}[x]$ has the property that $f(x+y)=f(x)+$ $f(y) \in \mathbb{Z} / p \mathbb{Z}[x, y]$. Show that $f(x)$ must be of the form $a_{0} x+a_{1} x^{p}+a_{2} x^{p^{2}}+$ $\cdots+a_{m} x^{p^{m}}$.
lean statement:

codex statement:
theorem exists_poly_of_additive_poly:
  fixes p::nat and f::"int poly"
  assumes "\<forall>x y. f (x + y) = f x + f y"
  shows "\<exists>a. f = (\<Sum>i<p. a i * X^(p^i))"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_7_24: undefined oops

(*
problem_number:12_12
natural language statement:
Show that $\sin (\pi / 12)$ is an algebraic number.
lean statement:
theorem exercise_12_12 : is_algebraic \<rat> (sin (real.pi/12)) :=

codex statement:
theorem algebraic_of_sin_pi_12:
  shows "algebraic (sin (\<pi> / 12))"
Our comment on the codex statement: unfortunately \<pi> is only a variable name
 *)
theorem exercise_12_12:
  shows "algebraic (sin (pi / 12))"
  oops

(*
problem_number:12_19
natural language statement:
Show that a finite integral domain is a field.
lean statement:

codex statement:
theorem finite_integral_domain_is_field:
  fixes R::"'a::comm_ring_1"
  assumes "finite (carrier R)" "\<forall>x\<in>carrier R. x \<noteq> 0 \<longrightarrow> \<exists>y\<in>carrier R. y * x = 1"
  shows "field R"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_12_19: undefined oops

(*
problem_number:12_22
natural language statement:
Let $F \subset E$ be algebraic number fields. Show that any isomorphism of $F$ into $\mathbb{C}$ extends in exactly $[E: F]$ ways to an isomorphism of $E$ into $\mathbb{C}$.
lean statement:

codex statement:
theorem exists_extension_of_isomorphism_of_algebraic_number_field:
  fixes F E::"'a::{field_char_0, algebraic_semidom}"
  assumes "F \<subseteq> E" "algebraic_semidom F" "algebraic_semidom E" "isomorphism F \<complex>"
  shows "\<exists>f. isomorphism E \<complex> \<and> \<forall>x\<in>F. f x = (isomorphism_to_complex F) x"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_12_22: undefined oops

(*
problem_number:12_30
natural language statement:
Let $p$ be an odd prime and consider $\mathbb{Q}(\sqrt{p})$. If $q \neq p$ is prime show that $\sigma_{q}(\sqrt{p})=$ $(p / q) \sqrt{p}$ where $\sigma_{q}$ is the Frobenius automorphism at a prime ideal in $\mathbb{Q}(\sqrt{p})$ lying above $q$.
lean statement:

codex statement:
theorem frobenius_automorphism_of_prime_ideal_of_sqrt_prime:
  fixes p q::nat
  assumes "prime p" "prime q" "q \<noteq> p"
  shows "frobenius_automorphism (prime_ideal_of_sqrt_prime p q) = (\<lambda>x. (p / q) * x)"
Our comment on the codex statement: <YOU CAN LEAVE YOUR COMMENT HERE>
 *)
theorem exercise_12_30: undefined oops

(*
problem_number:18_1
natural language statement:
Show that $165 x^{2}-21 y^{2}=19$ has no integral solution.
lean statement:
theorem exercise_18_1 : \<not> \<exists> x y : \<int>, 165*x^2 - 21*y^2 = 19 :=

codex statement:
theorem no_integral_solution_of_165_x_square_minus_21_y_square_eq_19:
  assumes "\<exists>x y::int. 165 * x^2 - 21 * y^2 = 19"
  shows False
Our comment on the codex statement:  the conclusion should be a negated existential as opposed to False
 *)
theorem exercise_18_1:
    shows "\<nexists>x y::int. 165 * x^2 - 21 * y^2 = 19"
oops

(*
problem_number:18_4
natural language statement:
Show that 1729 is the smallest positive integer expressible as the sum of two different integral cubes in two ways.
lean statement:
theorem exercise_18_4 {n : \<nat>} (hn : \<exists> x y z w : \<int>,
  x^3 + y^3 = n \<and> z^3 + w^3 = n \<and> x \<noteq> z \<and> x \<noteq> w \<and> y \<noteq> z \<and> y \<noteq> w) :
  n \<ge> 1729 :=

codex statement:
theorem smallest_positive_integer_expressible_as_sum_of_two_different_integral_cubes_in_two_ways:
  fixes a b c d::nat
  assumes "a^3 + b^3 = c^3 + d^3" "a^3 \<noteq> b^3" "c^3 \<noteq> d^3" "a^3 + b^3 = 1729"
  shows "\<forall>x y z w::nat. x^3 + y^3 = z^3 + w^3 \<longrightarrow> x^3 \<noteq> y^3 \<longrightarrow> z^3 \<noteq> w^3 \<longrightarrow> x^3 + y^3 \<ge> 1729"
Our comment on the codex statement:  there is only one way to write this
 *)
theorem exercise_18_4:
  shows "1729 = Inf {n::nat. \<exists>x y z w. x^3 + y^3 = n \<and> z^3 + w^3 = n \<and> x \<noteq> z \<and> x \<noteq> w \<and> y \<noteq> z \<and> y \<noteq> w}"
  oops

(*
problem_number:18_32
natural language statement:
Let $d$ be a square-free integer $d \equiv 1$ or 2 modulo 4 . Show that if $x$ and $y$ are integers such that $y^{2}=x^{3}-d$ then $(x, 2 d)=1$.
lean statement:

codex statement:
theorem gcd_eq_one_of_square_free_cong_one_or_two_mod_four:
  fixes d::int and x y::nat
  assumes "square_free d" "d \<equiv> 1 \<or> d \<equiv> 2 [MOD 4]" "y^2 = x^3 - d"
  shows "gcd x (2*d) = 1"
Our comment on the codex statement:  not bad except for the syntax of modular equivalence, and could use coprime.
 *)
theorem exercise_18_32:
  fixes d::int and x y::nat
  assumes "squarefree d" "[d=1] (mod 4) \<or> [d=2] (mod 4)" "y^2 = x^3 - d"
  shows "gcd x (2*d) = 1"
  oops

end