Theory Quadratic_Reciprocity
theory Quadratic_Reciprocity
imports Gauss
begin
text ‹
The proof is based on Gauss's fifth proof, which can be found at
🌐‹https://www.lehigh.edu/~shw2/q-recip/gauss5.pdf›.
›
locale QR =
fixes p :: "nat"
fixes q :: "nat"
assumes p_prime: "prime p"
assumes p_ge_2: "2 < p"
assumes q_prime: "prime q"
assumes q_ge_2: "2 < q"
assumes pq_neq: "p ≠ q"
begin
lemma odd_p: "odd p"
using p_ge_2 p_prime prime_odd_nat by blast
lemma p_ge_0: "0 < int p"
by (simp add: p_prime prime_gt_0_nat)
lemma p_eq2: "int p = (2 * ((int p - 1) div 2)) + 1"
using odd_p by simp
lemma odd_q: "odd q"
using q_ge_2 q_prime prime_odd_nat by blast
lemma q_ge_0: "0 < int q"
by (simp add: q_prime prime_gt_0_nat)
lemma q_eq2: "int q = (2 * ((int q - 1) div 2)) + 1"
using odd_q by simp
lemma pq_eq2: "int p * int q = (2 * ((int p * int q - 1) div 2)) + 1"
using odd_p odd_q by simp
lemma pq_coprime: "coprime p q"
using pq_neq p_prime primes_coprime_nat q_prime by blast
lemma pq_coprime_int: "coprime (int p) (int q)"
by (simp add: gcd_int_def pq_coprime)
lemma qp_ineq: "int p * k ≤ (int p * int q - 1) div 2 ⟷ k ≤ (int q - 1) div 2"
proof -
have "2 * int p * k ≤ int p * int q - 1 ⟷ 2 * k ≤ int q - 1"
using p_ge_0 by auto
then show ?thesis by auto
qed
lemma QRqp: "QR q p"
using QR_def QR_axioms by simp
lemma pq_commute: "int p * int q = int q * int p"
by simp
lemma pq_ge_0: "int p * int q > 0"
using p_ge_0 q_ge_0 mult_pos_pos by blast
definition "r = ((p - 1) div 2) * ((q - 1) div 2)"
definition "m = card (GAUSS.E p q)"
definition "n = card (GAUSS.E q p)"
abbreviation "Res k ≡ {0 .. k - 1}" for k :: int
abbreviation "Res_ge_0 k ≡ {0 <.. k - 1}" for k :: int
abbreviation "Res_0 k ≡ {0::int}" for k :: int
abbreviation "Res_l k ≡ {0 <.. (k - 1) div 2}" for k :: int
abbreviation "Res_h k ≡ {(k - 1) div 2 <.. k - 1}" for k :: int
abbreviation "Sets_pq r0 r1 r2 ≡
{(x::int). x ∈ r0 (int p * int q) ∧ x mod p ∈ r1 (int p) ∧ x mod q ∈ r2 (int q)}"
definition "A = Sets_pq Res_l Res_l Res_h"
definition "B = Sets_pq Res_l Res_h Res_l"
definition "C = Sets_pq Res_h Res_h Res_l"
definition "D = Sets_pq Res_l Res_h Res_h"
definition "E = Sets_pq Res_l Res_0 Res_h"
definition "F = Sets_pq Res_l Res_h Res_0"
definition "a = card A"
definition "b = card B"
definition "c = card C"
definition "d = card D"
definition "e = card E"
definition "f = card F"
lemma Gpq: "GAUSS p q"
using p_prime pq_neq p_ge_2 q_prime
by (auto simp: GAUSS_def cong_iff_dvd_diff dest: primes_dvd_imp_eq)
lemma Gqp: "GAUSS q p"
by (simp add: QRqp QR.Gpq)
lemma QR_lemma_01: "(λx. x mod q) ` E = GAUSS.E q p"
proof
have "x ∈ E ⟶ x mod int q ∈ GAUSS.E q p" if "x ∈ E" for x
proof -
from that obtain k where k: "x = int p * k"
unfolding E_def by blast
from that E_def have "x ∈ Res_l (int p * int q)"
by blast
then have "k ∈ GAUSS.A q"
using Gqp GAUSS.A_def k qp_ineq by (simp add: zero_less_mult_iff)
then have "x mod q ∈ GAUSS.E q p"
using GAUSS.C_def[of q p] Gqp k GAUSS.B_def[of q p] that GAUSS.E_def[of q p]
by (force simp: E_def)
then show ?thesis by auto
qed
then show "(λx. x mod int q) ` E ⊆ GAUSS.E q p"
by auto
show "GAUSS.E q p ⊆ (λx. x mod q) ` E"
proof
fix x
assume x: "x ∈ GAUSS.E q p"
then obtain ka where ka: "ka ∈ GAUSS.A q" "x = (ka * p) mod q"
by (auto simp: Gqp GAUSS.B_def GAUSS.C_def GAUSS.E_def)
then have "ka * p ∈ Res_l (int p * int q)"
using Gqp p_ge_0 qp_ineq by (simp add: GAUSS.A_def Groups.mult_ac(2))
then show "x ∈ (λx. x mod q) ` E"
using ka x Gqp q_ge_0 by (force simp: E_def GAUSS.E_def)
qed
qed
lemma QR_lemma_02: "e = n"
proof -
have "x = y" if x: "x ∈ E" and y: "y ∈ E" and mod: "x mod q = y mod q" for x y
proof -
obtain p_inv where p_inv: "[int p * p_inv = 1] (mod int q)"
using pq_coprime_int cong_solve_coprime_int by blast
from x y E_def obtain kx ky where k: "x = int p * kx" "y = int p * ky"
using dvd_def[of p x] by blast
with x y E_def have "0 < x" "int p * kx ≤ (int p * int q - 1) div 2"
"0 < y" "int p * ky ≤ (int p * int q - 1) div 2"
using greaterThanAtMost_iff mem_Collect_eq by blast+
with k have "0 ≤ kx" "kx < q" "0 ≤ ky" "ky < q"
using qp_ineq by (simp_all add: zero_less_mult_iff)
moreover from mod k have "(p_inv * (p * kx)) mod q = (p_inv * (p * ky)) mod q"
using mod_mult_cong by blast
then have "(p * p_inv * kx) mod q = (p * p_inv * ky) mod q"
by (simp add: algebra_simps)
then have "kx mod q = ky mod q"
using p_inv mod_mult_cong[of "p * p_inv" "q" "1"]
by (auto simp: cong_def)
then have "[kx = ky] (mod q)"
unfolding cong_def by blast
ultimately show ?thesis
using cong_less_imp_eq_int k by blast
qed
then have "inj_on (λx. x mod q) E"
by (auto simp: inj_on_def)
then show ?thesis
using QR_lemma_01 card_image e_def n_def by fastforce
qed
lemma QR_lemma_03: "f = m"
proof -
have "F = QR.E q p"
unfolding F_def pq_commute using QRqp QR.E_def[of q p] by fastforce
then have "f = QR.e q p"
unfolding f_def using QRqp QR.e_def[of q p] by presburger
then show ?thesis
using QRqp QR.QR_lemma_02 m_def QRqp QR.n_def by presburger
qed
definition f_1 :: "int ⇒ int × int"
where "f_1 x = ((x mod p), (x mod q))"
definition P_1 :: "int × int ⇒ int ⇒ bool"
where "P_1 res x ⟷ x mod p = fst res ∧ x mod q = snd res ∧ x ∈ Res (int p * int q)"
definition g_1 :: "int × int ⇒ int"
where "g_1 res = (THE x. P_1 res x)"
lemma P_1_lemma:
fixes res :: "int × int"
assumes "0 ≤ fst res" "fst res < p" "0 ≤ snd res" "snd res < q"
shows "∃!x. P_1 res x"
proof -
obtain y k1 k2 where yk: "y = nat (fst res) + k1 * p" "y = nat (snd res) + k2 * q"
using chinese_remainder[of p q] pq_coprime p_ge_0 q_ge_0 by fastforce
have "fst res = int (y - k1 * p)"
using ‹0 ≤ fst res› yk(1) by simp
moreover have "snd res = int (y - k2 * q)"
using ‹0 ≤ snd res› yk(2) by simp
ultimately have res: "res = (int (y - k1 * p), int (y - k2 * q))"
by (simp add: prod_eq_iff)
have y: "k1 * p ≤ y" "k2 * q ≤ y"
using yk by simp_all
from y have *: "[y = nat (fst res)] (mod p)" "[y = nat (snd res)] (mod q)"
by (auto simp add: res cong_le_nat intro: exI [of _ k1] exI [of _ k2])
from * have "(y mod (int p * int q)) mod int p = fst res" "(y mod (int p * int q)) mod int q = snd res"
using y apply (auto simp add: res of_nat_mult [symmetric] of_nat_mod [symmetric] mod_mod_cancel simp del: of_nat_mult)
apply (metis ‹fst res = int (y - k1 * p)› assms(1) assms(2) cong_def mod_pos_pos_trivial nat_int of_nat_mod)
apply (metis ‹snd res = int (y - k2 * q)› assms(3) assms(4) cong_def mod_pos_pos_trivial nat_int of_nat_mod)
done
then obtain x where "P_1 res x"
unfolding P_1_def
using Divides.pos_mod_bound Divides.pos_mod_sign pq_ge_0 by fastforce
moreover have "a = b" if "P_1 res a" "P_1 res b" for a b
proof -
from that have "int p * int q dvd a - b"
using divides_mult[of "int p" "a - b" "int q"] pq_coprime_int mod_eq_dvd_iff [of a _ b]
unfolding P_1_def by force
with that show ?thesis
using dvd_imp_le_int[of "a - b"] unfolding P_1_def by fastforce
qed
ultimately show ?thesis by auto
qed
lemma g_1_lemma:
fixes res :: "int × int"
assumes "0 ≤ fst res" "fst res < p" "0 ≤ snd res" "snd res < q"
shows "P_1 res (g_1 res)"
using assms P_1_lemma [of res] theI' [of "P_1 res"] g_1_def
by auto
definition "BuC = Sets_pq Res_ge_0 Res_h Res_l"
lemma finite_BuC [simp]:
"finite BuC"
proof -
{
fix p q :: nat
have "finite {x. 0 < x ∧ x < int p * int q}"
by simp
then have "finite {x.
0 < x ∧
x < int p * int q ∧
(int p - 1) div 2
< x mod int p ∧
x mod int p < int p ∧
0 < x mod int q ∧
x mod int q ≤ (int q - 1) div 2}"
by (auto intro: rev_finite_subset)
}
then show ?thesis
by (simp add: BuC_def)
qed
lemma QR_lemma_04: "card BuC = card (Res_h p × Res_l q)"
using card_bij_eq[of f_1 "BuC" "Res_h p × Res_l q" g_1]
proof
show "inj_on f_1 BuC"
proof
fix x y
assume *: "x ∈ BuC" "y ∈ BuC" "f_1 x = f_1 y"
then have "int p * int q dvd x - y"
using f_1_def pq_coprime_int divides_mult[of "int p" "x - y" "int q"]
mod_eq_dvd_iff[of x _ y]
by auto
with * show "x = y"
using dvd_imp_le_int[of "x - y" "int p * int q"] unfolding BuC_def by force
qed
show "inj_on g_1 (Res_h p × Res_l q)"
proof
fix x y
assume *: "x ∈ Res_h p × Res_l q" "y ∈ Res_h p × Res_l q" "g_1 x = g_1 y"
then have "0 ≤ fst x" "fst x < p" "0 ≤ snd x" "snd x < q"
"0 ≤ fst y" "fst y < p" "0 ≤ snd y" "snd y < q"
using mem_Sigma_iff prod.collapse by fastforce+
with * show "x = y"
using g_1_lemma[of x] g_1_lemma[of y] P_1_def by fastforce
qed
show "g_1 ` (Res_h p × Res_l q) ⊆ BuC"
proof
fix y
assume "y ∈ g_1 ` (Res_h p × Res_l q)"
then obtain x where x: "y = g_1 x" "x ∈ Res_h p × Res_l q"
by blast
then have "P_1 x y"
using g_1_lemma by fastforce
with x show "y ∈ BuC"
unfolding P_1_def BuC_def mem_Collect_eq using SigmaE prod.sel by fastforce
qed
qed (auto simp: finite_subset f_1_def, simp_all add: BuC_def)
lemma QR_lemma_05: "card (Res_h p × Res_l q) = r"
proof -
have "card (Res_l q) = (q - 1) div 2" "card (Res_h p) = (p - 1) div 2"
using p_eq2 by force+
then show ?thesis
unfolding r_def using card_cartesian_product[of "Res_h p" "Res_l q"] by presburger
qed
lemma QR_lemma_06: "b + c = r"
proof -
have "B ∩ C = {}" "finite B" "finite C" "B ∪ C = BuC"
unfolding B_def C_def BuC_def by fastforce+
then show ?thesis
unfolding b_def c_def using card.empty card_Un_Int QR_lemma_04 QR_lemma_05 by fastforce
qed
definition f_2:: "int ⇒ int"
where "f_2 x = (int p * int q) - x"
lemma f_2_lemma_1: "f_2 (f_2 x) = x"
by (simp add: f_2_def)
lemma f_2_lemma_2: "[f_2 x = int p - x] (mod p)"
by (simp add: f_2_def cong_iff_dvd_diff)
lemma f_2_lemma_3: "f_2 x ∈ S ⟹ x ∈ f_2 ` S"
using f_2_lemma_1[of x] image_eqI[of x f_2 "f_2 x" S] by presburger
lemma QR_lemma_07:
"f_2 ` Res_l (int p * int q) = Res_h (int p * int q)"
"f_2 ` Res_h (int p * int q) = Res_l (int p * int q)"
proof -
have 1: "f_2 ` Res_l (int p * int q) ⊆ Res_h (int p * int q)"
by (force simp: f_2_def)
have 2: "f_2 ` Res_h (int p * int q) ⊆ Res_l (int p * int q)"
using pq_eq2 by (fastforce simp: f_2_def)
from 2 have 3: "Res_h (int p * int q) ⊆ f_2 ` Res_l (int p * int q)"
using f_2_lemma_3 by blast
from 1 have 4: "Res_l (int p * int q) ⊆ f_2 ` Res_h (int p * int q)"
using f_2_lemma_3 by blast
from 1 3 show "f_2 ` Res_l (int p * int q) = Res_h (int p * int q)"
by blast
from 2 4 show "f_2 ` Res_h (int p * int q) = Res_l (int p * int q)"
by blast
qed
lemma QR_lemma_08:
"f_2 x mod p ∈ Res_l p ⟷ x mod p ∈ Res_h p"
"f_2 x mod p ∈ Res_h p ⟷ x mod p ∈ Res_l p"
using f_2_lemma_2[of x] cong_def[of "f_2 x" "p - x" p] minus_mod_self2[of x p]
zmod_zminus1_eq_if[of x p] p_eq2
by auto
lemma QR_lemma_09:
"f_2 x mod q ∈ Res_l q ⟷ x mod q ∈ Res_h q"
"f_2 x mod q ∈ Res_h q ⟷ x mod q ∈ Res_l q"
using QRqp QR.QR_lemma_08 f_2_def QR.f_2_def pq_commute by auto
lemma QR_lemma_10: "a = c"
unfolding a_def c_def
apply (rule card_bij_eq[of f_2 A C f_2])
unfolding A_def C_def
using QR_lemma_07 QR_lemma_08 QR_lemma_09 apply ((simp add: inj_on_def f_2_def), blast)+
apply fastforce+
done
definition "BuD = Sets_pq Res_l Res_h Res_ge_0"
definition "BuDuF = Sets_pq Res_l Res_h Res"
definition f_3 :: "int ⇒ int × int"
where "f_3 x = (x mod p, x div p + 1)"
definition g_3 :: "int × int ⇒ int"
where "g_3 x = fst x + (snd x - 1) * p"
lemma QR_lemma_11: "card BuDuF = card (Res_h p × Res_l q)"
using card_bij_eq[of f_3 BuDuF "Res_h p × Res_l q" g_3]
proof
show "f_3 ` BuDuF ⊆ Res_h p × Res_l q"
proof
fix y
assume "y ∈ f_3 ` BuDuF"
then obtain x where x: "y = f_3 x" "x ∈ BuDuF"
by blast
then have "x ≤ int p * (int q - 1) div 2 + (int p - 1) div 2"
unfolding BuDuF_def using p_eq2 int_distrib(4) by auto
moreover from x have "(int p - 1) div 2 ≤ - 1 + x mod p"
by (auto simp: BuDuF_def)
moreover have "int p * (int q - 1) div 2 = int p * ((int q - 1) div 2)"
by (subst div_mult1_eq) (simp add: odd_q)
then have "p * (int q - 1) div 2 = p * ((int q + 1) div 2 - 1)"
by fastforce
ultimately have "x ≤ p * ((int q + 1) div 2 - 1) - 1 + x mod p"
by linarith
then have "x div p < (int q + 1) div 2 - 1"
using mult.commute[of "int p" "x div p"] p_ge_0 div_mult_mod_eq[of x p]
and mult_less_cancel_left_pos[of p "x div p" "(int q + 1) div 2 - 1"]
by linarith
moreover from x have "0 < x div p + 1"
using pos_imp_zdiv_neg_iff[of p x] p_ge_0 by (auto simp: BuDuF_def)
ultimately show "y ∈ Res_h p × Res_l q"
using x by (auto simp: BuDuF_def f_3_def)
qed
show "inj_on g_3 (Res_h p × Res_l q)"
proof
have *: "f_3 (g_3 x) = x" if "x ∈ Res_h p × Res_l q" for x
proof -
from that have *: "(fst x + (snd x - 1) * int p) mod int p = fst x"
by force
from that have "(fst x + (snd x - 1) * int p) div int p + 1 = snd x"
by auto
with * show "f_3 (g_3 x) = x"
by (simp add: f_3_def g_3_def)
qed
fix x y
assume "x ∈ Res_h p × Res_l q" "y ∈ Res_h p × Res_l q" "g_3 x = g_3 y"
from this *[of x] *[of y] show "x = y"
by presburger
qed
show "g_3 ` (Res_h p × Res_l q) ⊆ BuDuF"
proof
fix y
assume "y ∈ g_3 ` (Res_h p × Res_l q)"
then obtain x where x: "x ∈ Res_h p × Res_l q" and y: "y = g_3 x"
by blast
then have "snd x ≤ (int q - 1) div 2"
by force
moreover have "int p * ((int q - 1) div 2) = (int p * int q - int p) div 2"
using int_distrib(4) div_mult1_eq[of "int p" "int q - 1" 2] odd_q by fastforce
ultimately have "(snd x) * int p ≤ (int q * int p - int p) div 2"
using mult_right_mono[of "snd x" "(int q - 1) div 2" p] mult.commute[of "(int q - 1) div 2" p]
pq_commute
by presburger
then have "(snd x - 1) * int p ≤ (int q * int p - 1) div 2 - int p"
using p_ge_0 int_distrib(3) by auto
moreover from x have "fst x ≤ int p - 1" by force
ultimately have "fst x + (snd x - 1) * int p ≤ (int p * int q - 1) div 2"
using pq_commute by linarith
moreover from x have "0 < fst x" "0 ≤ (snd x - 1) * p"
by fastforce+
ultimately show "y ∈ BuDuF"
unfolding BuDuF_def using q_ge_0 x g_3_def y by auto
qed
show "finite BuDuF" unfolding BuDuF_def by fastforce
qed (simp add: inj_on_inverseI[of BuDuF g_3] f_3_def g_3_def QR_lemma_05)+
lemma QR_lemma_12: "b + d + m = r"
proof -
have "B ∩ D = {}" "finite B" "finite D" "B ∪ D = BuD"
unfolding B_def D_def BuD_def by fastforce+
then have "b + d = card BuD"
unfolding b_def d_def using card_Un_Int by fastforce
moreover have "BuD ∩ F = {}" "finite BuD" "finite F"
unfolding BuD_def F_def by fastforce+
moreover have "BuD ∪ F = BuDuF"
unfolding BuD_def F_def BuDuF_def
using q_ge_0 ivl_disj_un_singleton(5)[of 0 "int q - 1"] by auto
ultimately show ?thesis
using QR_lemma_03 QR_lemma_05 QR_lemma_11 card_Un_disjoint[of BuD F]
unfolding b_def d_def f_def
by presburger
qed
lemma QR_lemma_13: "a + d + n = r"
proof -
have "A = QR.B q p"
unfolding A_def pq_commute using QRqp QR.B_def[of q p] by blast
then have "a = QR.b q p"
using a_def QRqp QR.b_def[of q p] by presburger
moreover have "D = QR.D q p"
unfolding D_def pq_commute using QRqp QR.D_def[of q p] by blast
then have "d = QR.d q p"
using d_def QRqp QR.d_def[of q p] by presburger
moreover have "n = QR.m q p"
using n_def QRqp QR.m_def[of q p] by presburger
moreover have "r = QR.r q p"
unfolding r_def using QRqp QR.r_def[of q p] by auto
ultimately show ?thesis
using QRqp QR.QR_lemma_12 by presburger
qed
lemma QR_lemma_14: "(-1::int) ^ (m + n) = (-1) ^ r"
proof -
have "m + n + 2 * d = r"
using QR_lemma_06 QR_lemma_10 QR_lemma_12 QR_lemma_13 by auto
then show ?thesis
using power_add[of "-1::int" "m + n" "2 * d"] by fastforce
qed
lemma Quadratic_Reciprocity:
"Legendre p q * Legendre q p = (-1::int) ^ ((p - 1) div 2 * ((q - 1) div 2))"
using Gpq Gqp GAUSS.gauss_lemma power_add[of "-1::int" m n] QR_lemma_14
unfolding r_def m_def n_def by auto
end
theorem Quadratic_Reciprocity:
assumes "prime p" "2 < p" "prime q" "2 < q" "p ≠ q"
shows "Legendre p q * Legendre q p = (-1::int) ^ ((p - 1) div 2 * ((q - 1) div 2))"
using QR.Quadratic_Reciprocity QR_def assms by blast
theorem Quadratic_Reciprocity_int:
assumes "prime (nat p)" "2 < p" "prime (nat q)" "2 < q" "p ≠ q"
shows "Legendre p q * Legendre q p = (-1::int) ^ (nat ((p - 1) div 2 * ((q - 1) div 2)))"
proof -
from assms have "0 ≤ (p - 1) div 2" by simp
moreover have "(nat p - 1) div 2 = nat ((p - 1) div 2)" "(nat q - 1) div 2 = nat ((q - 1) div 2)"
by fastforce+
ultimately have "(nat p - 1) div 2 * ((nat q - 1) div 2) = nat ((p - 1) div 2 * ((q - 1) div 2))"
using nat_mult_distrib by presburger
moreover have "2 < nat p" "2 < nat q" "nat p ≠ nat q" "int (nat p) = p" "int (nat q) = q"
using assms by linarith+
ultimately show ?thesis
using Quadratic_Reciprocity[of "nat p" "nat q"] assms by presburger
qed
end