P. 158 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 15701-15800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	lgslem2 15701	The set 𝑍 of all integers with absolute value at most 1 contains {-1, 0, 1}. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝑍 = {𝑥 ∈ ℤ ∣ (abs‘𝑥) ≤ 1}    ⇒   ⊢ (-1 ∈ 𝑍 ∧ 0 ∈ 𝑍 ∧ 1 ∈ 𝑍)
Theorem	lgslem3 15702*	The set 𝑍 of all integers with absolute value at most 1 is closed under multiplication. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝑍 = {𝑥 ∈ ℤ ∣ (abs‘𝑥) ≤ 1}    ⇒   ⊢ ((𝐴 ∈ 𝑍 ∧ 𝐵 ∈ 𝑍) → (𝐴 · 𝐵) ∈ 𝑍)
Theorem	lgslem4 15703*	Lemma for lgsfcl2 15706. (Contributed by Mario Carneiro, 4-Feb-2015.) (Proof shortened by AV, 19-Mar-2022.)
⊢ 𝑍 = {𝑥 ∈ ℤ ∣ (abs‘𝑥) ≤ 1}    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ((((𝐴↑((𝑃 − 1) / 2)) + 1) mod 𝑃) − 1) ∈ 𝑍)
Theorem	lgsval 15704*	Value of the Legendre symbol at an arbitrary integer. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L 𝑁) = if(𝑁 = 0, if((𝐴↑2) = 1, 1, 0), (if((𝑁 < 0 ∧ 𝐴 < 0), -1, 1) · (seq1( · , 𝐹)‘(abs‘𝑁)))))
Theorem	lgsfvalg 15705*	Value of the function 𝐹 which defines the Legendre symbol at the primes. (Contributed by Mario Carneiro, 4-Feb-2015.) (Revised by Jim Kingdon, 4-Nov-2024.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ 𝑀 ∈ ℕ) → (𝐹‘𝑀) = if(𝑀 ∈ ℙ, (if(𝑀 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑀 − 1) / 2)) + 1) mod 𝑀) − 1))↑(𝑀 pCnt 𝑁)), 1))
Theorem	lgsfcl2 15706*	The function 𝐹 is closed in integers with absolute value less than 1 (namely {-1, 0, 1}, see zabsle1 15699). (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    &   ⊢ 𝑍 = {𝑥 ∈ ℤ ∣ (abs‘𝑥) ≤ 1}    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → 𝐹:ℕ⟶𝑍)
Theorem	lgscllem 15707*	The Legendre symbol is an element of 𝑍. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    &   ⊢ 𝑍 = {𝑥 ∈ ℤ ∣ (abs‘𝑥) ≤ 1}    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L 𝑁) ∈ 𝑍)
Theorem	lgsfcl 15708*	Closure of the function 𝐹 which defines the Legendre symbol at the primes. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → 𝐹:ℕ⟶ℤ)
Theorem	lgsfle1 15709*	The function 𝐹 has magnitude less or equal to 1. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ (((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) ∧ 𝑀 ∈ ℕ) → (abs‘(𝐹‘𝑀)) ≤ 1)
Theorem	lgsval2lem 15710*	Lemma for lgsval2 15716. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℙ) → (𝐴 /L 𝑁) = if(𝑁 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑁 − 1) / 2)) + 1) mod 𝑁) − 1)))
Theorem	lgsval4lem 15711*	Lemma for lgsval4 15720. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (if(𝑛 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑛 − 1) / 2)) + 1) mod 𝑛) − 1))↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, ((𝐴 /L 𝑛)↑(𝑛 pCnt 𝑁)), 1)))
Theorem	lgscl2 15712*	The Legendre symbol is an integer with absolute value less than or equal to 1. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝑍 = {𝑥 ∈ ℤ ∣ (abs‘𝑥) ≤ 1}    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L 𝑁) ∈ 𝑍)
Theorem	lgs0 15713	The Legendre symbol when the second argument is zero. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (𝐴 ∈ ℤ → (𝐴 /L 0) = if((𝐴↑2) = 1, 1, 0))
Theorem	lgscl 15714	The Legendre symbol is an integer. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L 𝑁) ∈ ℤ)
Theorem	lgsle1 15715	The Legendre symbol has absolute value less than or equal to 1. Together with lgscl 15714 this implies that it takes values in {-1, 0, 1}. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (abs‘(𝐴 /L 𝑁)) ≤ 1)
Theorem	lgsval2 15716	The Legendre symbol at a prime (this is the traditional domain of the Legendre symbol, except for the addition of prime 2). (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑃 ∈ ℙ) → (𝐴 /L 𝑃) = if(𝑃 = 2, if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)), ((((𝐴↑((𝑃 − 1) / 2)) + 1) mod 𝑃) − 1)))
Theorem	lgs2 15717	The Legendre symbol at 2. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (𝐴 ∈ ℤ → (𝐴 /L 2) = if(2 ∥ 𝐴, 0, if((𝐴 mod 8) ∈ {1, 7}, 1, -1)))
Theorem	lgsval3 15718	The Legendre symbol at an odd prime (this is the traditional domain of the Legendre symbol). (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → (𝐴 /L 𝑃) = ((((𝐴↑((𝑃 − 1) / 2)) + 1) mod 𝑃) − 1))
Theorem	lgsvalmod 15719	The Legendre symbol is equivalent to 𝑎↑((𝑝 − 1) / 2), mod 𝑝. This theorem is also called "Euler's criterion", see theorem 9.2 in [ApostolNT] p. 180, or a representation of Euler's criterion using the Legendre symbol, (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ((𝐴 /L 𝑃) mod 𝑃) = ((𝐴↑((𝑃 − 1) / 2)) mod 𝑃))
Theorem	lgsval4 15720*	Restate lgsval 15704 for nonzero 𝑁, where the function 𝐹 has been abbreviated into a self-referential expression taking the value of /L on the primes as given. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, ((𝐴 /L 𝑛)↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝐴 /L 𝑁) = (if((𝑁 < 0 ∧ 𝐴 < 0), -1, 1) · (seq1( · , 𝐹)‘(abs‘𝑁))))
Theorem	lgsfcl3 15721*	Closure of the function 𝐹 which defines the Legendre symbol at the primes. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, ((𝐴 /L 𝑛)↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → 𝐹:ℕ⟶ℤ)
Theorem	lgsval4a 15722*	Same as lgsval4 15720 for positive 𝑁. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, ((𝐴 /L 𝑛)↑(𝑛 pCnt 𝑁)), 1))    ⇒   ⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝐴 /L 𝑁) = (seq1( · , 𝐹)‘𝑁))
Theorem	lgscl1 15723	The value of the Legendre symbol is either -1 or 0 or 1. (Contributed by AV, 13-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L 𝑁) ∈ {-1, 0, 1})
Theorem	lgsneg 15724	The Legendre symbol is either even or odd under negation with respect to the second parameter according to the sign of the first. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝐴 /L -𝑁) = (if(𝐴 < 0, -1, 1) · (𝐴 /L 𝑁)))
Theorem	lgsneg1 15725	The Legendre symbol for nonnegative first parameter is unchanged by negation of the second. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝑁 ∈ ℤ) → (𝐴 /L -𝑁) = (𝐴 /L 𝑁))
Theorem	lgsmod 15726	The Legendre (Jacobi) symbol is preserved under reduction mod 𝑛 when 𝑛 is odd. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) → ((𝐴 mod 𝑁) /L 𝑁) = (𝐴 /L 𝑁))
Theorem	lgsdilem 15727	Lemma for lgsdi 15737 and lgsdir 15735: the sign part of the Legendre symbol is multiplicative. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → if((𝑁 < 0 ∧ (𝐴 · 𝐵) < 0), -1, 1) = (if((𝑁 < 0 ∧ 𝐴 < 0), -1, 1) · if((𝑁 < 0 ∧ 𝐵 < 0), -1, 1)))
Theorem	lgsdir2lem1 15728	Lemma for lgsdir2 15733. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (((1 mod 8) = 1 ∧ (-1 mod 8) = 7) ∧ ((3 mod 8) = 3 ∧ (-3 mod 8) = 5))
Theorem	lgsdir2lem2 15729	Lemma for lgsdir2 15733. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (𝐾 ∈ ℤ ∧ 2 ∥ (𝐾 + 1) ∧ ((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) → ((𝐴 mod 8) ∈ (0...𝐾) → (𝐴 mod 8) ∈ 𝑆)))    &   ⊢ 𝑀 = (𝐾 + 1)    &   ⊢ 𝑁 = (𝑀 + 1)    &   ⊢ 𝑁 ∈ 𝑆    ⇒   ⊢ (𝑁 ∈ ℤ ∧ 2 ∥ (𝑁 + 1) ∧ ((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) → ((𝐴 mod 8) ∈ (0...𝑁) → (𝐴 mod 8) ∈ 𝑆)))
Theorem	lgsdir2lem3 15730	Lemma for lgsdir2 15733. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) → (𝐴 mod 8) ∈ ({1, 7} ∪ {3, 5}))
Theorem	lgsdir2lem4 15731	Lemma for lgsdir2 15733. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝐴 mod 8) ∈ {1, 7}) → (((𝐴 · 𝐵) mod 8) ∈ {1, 7} ↔ (𝐵 mod 8) ∈ {1, 7}))
Theorem	lgsdir2lem5 15732	Lemma for lgsdir2 15733. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ ((𝐴 mod 8) ∈ {3, 5} ∧ (𝐵 mod 8) ∈ {3, 5})) → ((𝐴 · 𝐵) mod 8) ∈ {1, 7})
Theorem	lgsdir2 15733	The Legendre symbol is completely multiplicative at 2. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((𝐴 · 𝐵) /L 2) = ((𝐴 /L 2) · (𝐵 /L 2)))
Theorem	lgsdirprm 15734	The Legendre symbol is completely multiplicative at the primes. See theorem 9.3 in [ApostolNT] p. 180. (Contributed by Mario Carneiro, 4-Feb-2015.) (Proof shortened by AV, 18-Mar-2022.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑃 ∈ ℙ) → ((𝐴 · 𝐵) /L 𝑃) = ((𝐴 /L 𝑃) · (𝐵 /L 𝑃)))
Theorem	lgsdir 15735	The Legendre symbol is completely multiplicative in its left argument. Generalization of theorem 9.9(a) in [ApostolNT] p. 188 (which assumes that 𝐴 and 𝐵 are odd positive integers). (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0)) → ((𝐴 · 𝐵) /L 𝑁) = ((𝐴 /L 𝑁) · (𝐵 /L 𝑁)))
Theorem	lgsdilem2 15736*	Lemma for lgsdi 15737. (Contributed by Mario Carneiro, 4-Feb-2015.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝑁 ≠ 0)    &   ⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, ((𝐴 /L 𝑛)↑(𝑛 pCnt 𝑀)), 1))    ⇒   ⊢ (𝜑 → (seq1( · , 𝐹)‘(abs‘𝑀)) = (seq1( · , 𝐹)‘(abs‘(𝑀 · 𝑁))))
Theorem	lgsdi 15737	The Legendre symbol is completely multiplicative in its right argument. Generalization of theorem 9.9(b) in [ApostolNT] p. 188 (which assumes that 𝑀 and 𝑁 are odd positive integers). (Contributed by Mario Carneiro, 5-Feb-2015.)
⊢ (((𝐴 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝑀 ≠ 0 ∧ 𝑁 ≠ 0)) → (𝐴 /L (𝑀 · 𝑁)) = ((𝐴 /L 𝑀) · (𝐴 /L 𝑁)))
Theorem	lgsne0 15738	The Legendre symbol is nonzero (and hence equal to 1 or -1) precisely when the arguments are coprime. (Contributed by Mario Carneiro, 5-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐴 /L 𝑁) ≠ 0 ↔ (𝐴 gcd 𝑁) = 1))
Theorem	lgsabs1 15739	The Legendre symbol is nonzero (and hence equal to 1 or -1) precisely when the arguments are coprime. (Contributed by Mario Carneiro, 5-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘(𝐴 /L 𝑁)) = 1 ↔ (𝐴 gcd 𝑁) = 1))
Theorem	lgssq 15740	The Legendre symbol at a square is equal to 1. Together with lgsmod 15726 this implies that the Legendre symbol takes value 1 at every quadratic residue. (Contributed by Mario Carneiro, 5-Feb-2015.) (Revised by AV, 20-Jul-2021.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((𝐴↑2) /L 𝑁) = 1)
Theorem	lgssq2 15741	The Legendre symbol at a square is equal to 1. (Contributed by Mario Carneiro, 5-Feb-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ (𝐴 gcd 𝑁) = 1) → (𝐴 /L (𝑁↑2)) = 1)
Theorem	lgsprme0 15742	The Legendre symbol at any prime (even at 2) is 0 iff the prime does not divide the first argument. See definition in [ApostolNT] p. 179. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑃 ∈ ℙ) → ((𝐴 /L 𝑃) = 0 ↔ (𝐴 mod 𝑃) = 0))
Theorem	1lgs 15743	The Legendre symbol at 1. See example 1 in [ApostolNT] p. 180. (Contributed by Mario Carneiro, 28-Apr-2016.)
⊢ (𝑁 ∈ ℤ → (1 /L 𝑁) = 1)
Theorem	lgs1 15744	The Legendre symbol at 1. See definition in [ApostolNT] p. 188. (Contributed by Mario Carneiro, 28-Apr-2016.)
⊢ (𝐴 ∈ ℤ → (𝐴 /L 1) = 1)
Theorem	lgsmodeq 15745	The Legendre (Jacobi) symbol is preserved under reduction mod 𝑛 when 𝑛 is odd. Theorem 9.9(c) in [ApostolNT] p. 188. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁)) → ((𝐴 mod 𝑁) = (𝐵 mod 𝑁) → (𝐴 /L 𝑁) = (𝐵 /L 𝑁)))
Theorem	lgsmulsqcoprm 15746	The Legendre (Jacobi) symbol is preserved under multiplication with a square of an integer coprime to the second argument. Theorem 9.9(d) in [ApostolNT] p. 188. (Contributed by AV, 20-Jul-2021.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) ∧ (𝑁 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1)) → (((𝐴↑2) · 𝐵) /L 𝑁) = (𝐵 /L 𝑁))
Theorem	lgsdirnn0 15747	Variation on lgsdir 15735 valid for all 𝐴, 𝐵 but only for positive 𝑁. (The exact location of the failure of this law is for 𝐴 = 0, 𝐵 < 0, 𝑁 = -1 in which case (0 /L -1) = 1 but (𝐵 /L -1) = -1.) (Contributed by Mario Carneiro, 28-Apr-2016.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((𝐴 · 𝐵) /L 𝑁) = ((𝐴 /L 𝑁) · (𝐵 /L 𝑁)))
Theorem	lgsdinn0 15748	Variation on lgsdi 15737 valid for all 𝑀, 𝑁 but only for positive 𝐴. (The exact location of the failure of this law is for 𝐴 = -1, 𝑀 = 0, and some 𝑁 in which case (-1 /L 0) = 1 but (-1 /L 𝑁) = -1 when -1 is not a quadratic residue mod 𝑁.) (Contributed by Mario Carneiro, 28-Apr-2016.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L (𝑀 · 𝑁)) = ((𝐴 /L 𝑀) · (𝐴 /L 𝑁)))
11.3.5  Gauss' Lemma Gauss' Lemma is valid for any integer not dividing the given prime number. In the following, only the special case for 2 (not dividing any odd prime) is proven, see gausslemma2d 15769. The general case is still to prove.
Theorem	gausslemma2dlem0a 15749	Auxiliary lemma 1 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    ⇒   ⊢ (𝜑 → 𝑃 ∈ ℕ)
Theorem	gausslemma2dlem0b 15750	Auxiliary lemma 2 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝐻 ∈ ℕ)
Theorem	gausslemma2dlem0c 15751	Auxiliary lemma 3 for gausslemma2d 15769. (Contributed by AV, 13-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → ((!‘𝐻) gcd 𝑃) = 1)
Theorem	gausslemma2dlem0d 15752	Auxiliary lemma 4 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → 𝑀 ∈ ℕ0)
Theorem	gausslemma2dlem0e 15753	Auxiliary lemma 5 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (𝑀 · 2) < (𝑃 / 2))
Theorem	gausslemma2dlem0f 15754	Auxiliary lemma 6 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → (𝑀 + 1) ≤ 𝐻)
Theorem	gausslemma2dlem0g 15755	Auxiliary lemma 7 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝑀 ≤ 𝐻)
Theorem	gausslemma2dlem0h 15756	Auxiliary lemma 8 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → 𝑁 ∈ ℕ0)
Theorem	gausslemma2dlem0i 15757	Auxiliary lemma 9 for gausslemma2d 15769. (Contributed by AV, 14-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (((2 /L 𝑃) mod 𝑃) = ((-1↑𝑁) mod 𝑃) → (2 /L 𝑃) = (-1↑𝑁)))
Theorem	gausslemma2dlem1a 15758*	Lemma for gausslemma2dlem1 15761. (Contributed by AV, 1-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    ⇒   ⊢ (𝜑 → ran 𝑅 = (1...𝐻))
Theorem	gausslemma2dlem1cl 15759	Lemma for gausslemma2dlem1 15761. Closure of the body of the definition of 𝑅. (Contributed by Jim Kingdon, 10-Aug-2025.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ (𝜑 → 𝐴 ∈ (1...𝐻))    ⇒   ⊢ (𝜑 → if((𝐴 · 2) < (𝑃 / 2), (𝐴 · 2), (𝑃 − (𝐴 · 2))) ∈ ℤ)
Theorem	gausslemma2dlem1f1o 15760*	Lemma for gausslemma2dlem1 15761. (Contributed by Jim Kingdon, 9-Aug-2025.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    ⇒   ⊢ (𝜑 → 𝑅:(1...𝐻)–1-1-onto→(1...𝐻))
Theorem	gausslemma2dlem1 15761*	Lemma 1 for gausslemma2d 15769. (Contributed by AV, 5-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    ⇒   ⊢ (𝜑 → (!‘𝐻) = ∏𝑘 ∈ (1...𝐻)(𝑅‘𝑘))
Theorem	gausslemma2dlem2 15762*	Lemma 2 for gausslemma2d 15769. (Contributed by AV, 4-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → ∀𝑘 ∈ (1...𝑀)(𝑅‘𝑘) = (𝑘 · 2))
Theorem	gausslemma2dlem3 15763*	Lemma 3 for gausslemma2d 15769. (Contributed by AV, 4-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → ∀𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘) = (𝑃 − (𝑘 · 2)))
Theorem	gausslemma2dlem4 15764*	Lemma 4 for gausslemma2d 15769. (Contributed by AV, 16-Jun-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (!‘𝐻) = (∏𝑘 ∈ (1...𝑀)(𝑅‘𝑘) · ∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘)))
Theorem	gausslemma2dlem5a 15765*	Lemma for gausslemma2dlem5 15766. (Contributed by AV, 8-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘) mod 𝑃) = (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(-1 · (𝑘 · 2)) mod 𝑃))
Theorem	gausslemma2dlem5 15766*	Lemma 5 for gausslemma2d 15769. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘) mod 𝑃) = (((-1↑𝑁) · ∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑘 · 2)) mod 𝑃))
Theorem	gausslemma2dlem6 15767*	Lemma 6 for gausslemma2d 15769. (Contributed by AV, 16-Jun-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → ((!‘𝐻) mod 𝑃) = ((((-1↑𝑁) · (2↑𝐻)) · (!‘𝐻)) mod 𝑃))
Theorem	gausslemma2dlem7 15768*	Lemma 7 for gausslemma2d 15769. (Contributed by AV, 13-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (((-1↑𝑁) · (2↑𝐻)) mod 𝑃) = 1)
Theorem	gausslemma2d 15769*	Gauss' Lemma (see also theorem 9.6 in [ApostolNT] p. 182) for integer 2: Let p be an odd prime. Let S = {2, 4, 6, ..., p - 1}. Let n denote the number of elements of S whose least positive residue modulo p is greater than p/2. Then ( 2 | p ) = (-1)^n. (Contributed by AV, 14-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (2 /L 𝑃) = (-1↑𝑁))
11.3.6  Quadratic reciprocity
Theorem	lgseisenlem1 15770*	Lemma for lgseisen 15774. If 𝑅(𝑢) = (𝑄 · 𝑢) mod 𝑃 and 𝑀(𝑢) = (-1↑𝑅(𝑢)) · 𝑅(𝑢), then for any even 1 ≤ 𝑢 ≤ 𝑃 − 1, 𝑀(𝑢) is also an even integer 1 ≤ 𝑀(𝑢) ≤ 𝑃 − 1. To simplify these statements, we divide all the even numbers by 2, so that it becomes the statement that 𝑀(𝑥 / 2) = (-1↑𝑅(𝑥 / 2)) · 𝑅(𝑥 / 2) / 2 is an integer between 1 and (𝑃 − 1) / 2. (Contributed by Mario Carneiro, 17-Jun-2015.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    ⇒   ⊢ (𝜑 → 𝑀:(1...((𝑃 − 1) / 2))⟶(1...((𝑃 − 1) / 2)))
Theorem	lgseisenlem2 15771*	Lemma for lgseisen 15774. The function 𝑀 is an injection (and hence a bijection by the pigeonhole principle). (Contributed by Mario Carneiro, 17-Jun-2015.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   ⊢ 𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)    ⇒   ⊢ (𝜑 → 𝑀:(1...((𝑃 − 1) / 2))–1-1-onto→(1...((𝑃 − 1) / 2)))
Theorem	lgseisenlem3 15772*	Lemma for lgseisen 15774. (Contributed by Mario Carneiro, 17-Jun-2015.) (Proof shortened by AV, 28-Jul-2019.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   ⊢ 𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)    &   ⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝐺 = (mulGrp‘𝑌)    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    ⇒   ⊢ (𝜑 → (𝐺 Σg (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘((-1↑𝑅) · 𝑄)))) = (1r‘𝑌))
Theorem	lgseisenlem4 15773*	Lemma for lgseisen 15774. (Contributed by Mario Carneiro, 18-Jun-2015.) (Proof shortened by AV, 15-Jun-2019.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   ⊢ 𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)    &   ⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝐺 = (mulGrp‘𝑌)    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    ⇒   ⊢ (𝜑 → ((𝑄↑((𝑃 − 1) / 2)) mod 𝑃) = ((-1↑Σ𝑥 ∈ (1...((𝑃 − 1) / 2))(⌊‘((𝑄 / 𝑃) · (2 · 𝑥)))) mod 𝑃))
Theorem	lgseisen 15774*	Eisenstein's lemma, an expression for (𝑃 /L 𝑄) when 𝑃, 𝑄 are distinct odd primes. (Contributed by Mario Carneiro, 18-Jun-2015.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    ⇒   ⊢ (𝜑 → (𝑄 /L 𝑃) = (-1↑Σ𝑥 ∈ (1...((𝑃 − 1) / 2))(⌊‘((𝑄 / 𝑃) · (2 · 𝑥)))))
Theorem	lgsquadlemsfi 15775*	Lemma for lgsquad 15780. 𝑆 is finite. (Contributed by Jim Kingdon, 16-Sep-2025.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → 𝑆 ∈ Fin)
Theorem	lgsquadlemofi 15776*	Lemma for lgsquad 15780. There are finitely many members of 𝑆 with odd first part. (Contributed by Jim Kingdon, 16-Sep-2025.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → {𝑧 ∈ 𝑆 ∣ ¬ 2 ∥ (1st ‘𝑧)} ∈ Fin)
Theorem	lgsquadlem1 15777*	Lemma for lgsquad 15780. Count the members of 𝑆 with odd coordinates. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → (-1↑Σ𝑢 ∈ (((⌊‘(𝑀 / 2)) + 1)...𝑀)(⌊‘((𝑄 / 𝑃) · (2 · 𝑢)))) = (-1↑(♯‘{𝑧 ∈ 𝑆 ∣ ¬ 2 ∥ (1st ‘𝑧)})))
Theorem	lgsquadlem2 15778*	Lemma for lgsquad 15780. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 15777 to get the total count of lattice points in 𝑆 (up to parity). (Contributed by Mario Carneiro, 18-Jun-2015.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → (𝑄 /L 𝑃) = (-1↑(♯‘𝑆)))
Theorem	lgsquadlem3 15779*	Lemma for lgsquad 15780. (Contributed by Mario Carneiro, 18-Jun-2015.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → ((𝑃 /L 𝑄) · (𝑄 /L 𝑃)) = (-1↑(𝑀 · 𝑁)))
Theorem	lgsquad 15780	The Law of Quadratic Reciprocity, see also theorem 9.8 in [ApostolNT] p. 185. If 𝑃 and 𝑄 are distinct odd primes, then the product of the Legendre symbols (𝑃 /L 𝑄) and (𝑄 /L 𝑃) is the parity of ((𝑃 − 1) / 2) · ((𝑄 − 1) / 2). This uses Eisenstein's proof, which also has a nice geometric interpretation - see https://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity. This is Metamath 100 proof #7. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑄 ∈ (ℙ ∖ {2}) ∧ 𝑃 ≠ 𝑄) → ((𝑃 /L 𝑄) · (𝑄 /L 𝑃)) = (-1↑(((𝑃 − 1) / 2) · ((𝑄 − 1) / 2))))
Theorem	lgsquad2lem1 15781	Lemma for lgsquad2 15783. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    &   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 1)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → (𝐴 · 𝐵) = 𝑀)    &   ⊢ (𝜑 → ((𝐴 /L 𝑁) · (𝑁 /L 𝐴)) = (-1↑(((𝐴 − 1) / 2) · ((𝑁 − 1) / 2))))    &   ⊢ (𝜑 → ((𝐵 /L 𝑁) · (𝑁 /L 𝐵)) = (-1↑(((𝐵 − 1) / 2) · ((𝑁 − 1) / 2))))    ⇒   ⊢ (𝜑 → ((𝑀 /L 𝑁) · (𝑁 /L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))
Theorem	lgsquad2lem2 15782*	Lemma for lgsquad2 15783. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    &   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 1)    &   ⊢ ((𝜑 ∧ (𝑚 ∈ (ℙ ∖ {2}) ∧ (𝑚 gcd 𝑁) = 1)) → ((𝑚 /L 𝑁) · (𝑁 /L 𝑚)) = (-1↑(((𝑚 − 1) / 2) · ((𝑁 − 1) / 2))))    &   ⊢ (𝜓 ↔ ∀𝑥 ∈ (1...𝑘)((𝑥 gcd (2 · 𝑁)) = 1 → ((𝑥 /L 𝑁) · (𝑁 /L 𝑥)) = (-1↑(((𝑥 − 1) / 2) · ((𝑁 − 1) / 2)))))    ⇒   ⊢ (𝜑 → ((𝑀 /L 𝑁) · (𝑁 /L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))
Theorem	lgsquad2 15783	Extend lgsquad 15780 to coprime odd integers (the domain of the Jacobi symbol). (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    &   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 1)    ⇒   ⊢ (𝜑 → ((𝑀 /L 𝑁) · (𝑁 /L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))
Theorem	lgsquad3 15784	Extend lgsquad2 15783 to integers which share a factor. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ (((𝑀 ∈ ℕ ∧ ¬ 2 ∥ 𝑀) ∧ (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁)) → (𝑀 /L 𝑁) = ((-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))) · (𝑁 /L 𝑀)))
Theorem	m1lgs 15785	The first supplement to the law of quadratic reciprocity. Negative one is a square mod an odd prime 𝑃 iff 𝑃≡1 (mod 4). See first case of theorem 9.4 in [ApostolNT] p. 181. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ (𝑃 ∈ (ℙ ∖ {2}) → ((-1 /L 𝑃) = 1 ↔ (𝑃 mod 4) = 1))
Theorem	2lgslem1a1 15786*	Lemma 1 for 2lgslem1a 15788. (Contributed by AV, 16-Jun-2021.)
⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))
Theorem	2lgslem1a2 15787	Lemma 2 for 2lgslem1a 15788. (Contributed by AV, 18-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))
Theorem	2lgslem1a 15788*	Lemma 1 for 2lgslem1 15791. (Contributed by AV, 18-Jun-2021.)
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → {𝑥 ∈ ℤ ∣ ∃𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑥 = (𝑖 · 2) ∧ (𝑃 / 2) < (𝑥 mod 𝑃))} = {𝑥 ∈ ℤ ∣ ∃𝑖 ∈ (((⌊‘(𝑃 / 4)) + 1)...((𝑃 − 1) / 2))𝑥 = (𝑖 · 2)})
Theorem	2lgslem1b 15789*	Lemma 2 for 2lgslem1 15791. (Contributed by AV, 18-Jun-2021.)
⊢ 𝐼 = (𝐴...𝐵)    &   ⊢ 𝐹 = (𝑗 ∈ 𝐼 ↦ (𝑗 · 2))    ⇒   ⊢ 𝐹:𝐼–1-1-onto→{𝑥 ∈ ℤ ∣ ∃𝑖 ∈ 𝐼 𝑥 = (𝑖 · 2)}
Theorem	2lgslem1c 15790	Lemma 3 for 2lgslem1 15791. (Contributed by AV, 19-Jun-2021.)
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))
Theorem	2lgslem1 15791*	Lemma 1 for 2lgs 15804. (Contributed by AV, 19-Jun-2021.)
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (♯‘{𝑥 ∈ ℤ ∣ ∃𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑥 = (𝑖 · 2) ∧ (𝑃 / 2) < (𝑥 mod 𝑃))}) = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))))
Theorem	2lgslem2 15792	Lemma 2 for 2lgs 15804. (Contributed by AV, 20-Jun-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)
Theorem	2lgslem3a 15793	Lemma for 2lgslem3a1 15797. (Contributed by AV, 14-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))
Theorem	2lgslem3b 15794	Lemma for 2lgslem3b1 15798. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3c 15795	Lemma for 2lgslem3c1 15799. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3d 15796	Lemma for 2lgslem3d1 15800. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2))
Theorem	2lgslem3a1 15797	Lemma 1 for 2lgslem3 15801. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0)
Theorem	2lgslem3b1 15798	Lemma 2 for 2lgslem3 15801. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1)
Theorem	2lgslem3c1 15799	Lemma 3 for 2lgslem3 15801. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1)
Theorem	2lgslem3d1 15800	Lemma 4 for 2lgslem3 15801. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0)