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Theorem List for Metamath Proof Explorer - 25801-25900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremlgslem1 25801 When 𝑎 is coprime to the prime 𝑝, 𝑎↑((𝑝 − 1) / 2) is equivalent mod 𝑝 to 1 or -1, and so adding 1 makes it equivalent to 0 or 2. (Contributed by Mario Carneiro, 4-Feb-2015.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2}) ∧ ¬ 𝑃𝐴) → (((𝐴↑((𝑃 − 1) / 2)) + 1) mod 𝑃) ∈ {0, 2})
 
Theoremlgslem2 25802 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)
 
Theoremlgslem3 25803* 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}       ((𝐴𝑍𝐵𝑍) → (𝐴 · 𝐵) ∈ 𝑍)
 
Theoremlgslem4 25804* Lemma for lgsfcl2 25807. (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)
 
Theoremlgsval 25805* 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)))))
 
Theoremlgsfval 25806* Value 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))       (𝑀 ∈ ℕ → (𝐹𝑀) = 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))
 
Theoremlgsfcl2 25807* The function 𝐹 is closed in integers with absolute value less than 1 (namely {-1, 0, 1}, see zabsle1 25800). (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)
 
Theoremlgscllem 25808* 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)
 
Theoremlgsfcl 25809* 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)
 
Theoremlgsfle1 25810* 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‘(𝐹𝑀)) ≤ 1)
 
Theoremlgsval2lem 25811* Lemma for lgsval2 25817. (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)))
 
Theoremlgsval4lem 25812* Lemma for lgsval4 25821. (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)))
 
Theoremlgscl2 25813* 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)
 
Theoremlgs0 25814 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))
 
Theoremlgscl 25815 The Legendre symbol is an integer. (Contributed by Mario Carneiro, 4-Feb-2015.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L 𝑁) ∈ ℤ)
 
Theoremlgsle1 25816 The Legendre symbol has absolute value less than or equal to 1. Together with lgscl 25815 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)
 
Theoremlgsval2 25817 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)))
 
Theoremlgs2 25818 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)))
 
Theoremlgsval3 25819 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))
 
Theoremlgsvalmod 25820 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, see also lgsqr 25855. (Contributed by Mario Carneiro, 4-Feb-2015.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ((𝐴 /L 𝑃) mod 𝑃) = ((𝐴↑((𝑃 − 1) / 2)) mod 𝑃))
 
Theoremlgsval4 25821* Restate lgsval 25805 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))))
 
Theoremlgsfcl3 25822* 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)
 
Theoremlgsval4a 25823* Same as lgsval4 25821 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))
 
Theoremlgscl1 25824 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})
 
Theoremlgsneg 25825 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)))
 
Theoremlgsneg1 25826 The Legendre symbol for nonnegative first parameter is unchanged by negation of the second. (Contributed by Mario Carneiro, 4-Feb-2015.)
((𝐴 ∈ ℕ0𝑁 ∈ ℤ) → (𝐴 /L -𝑁) = (𝐴 /L 𝑁))
 
Theoremlgsmod 25827 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))
 
Theoremlgsdilem 25828 Lemma for lgsdi 25838 and lgsdir 25836: 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)))
 
Theoremlgsdir2lem1 25829 Lemma for lgsdir2 25834. (Contributed by Mario Carneiro, 4-Feb-2015.)
(((1 mod 8) = 1 ∧ (-1 mod 8) = 7) ∧ ((3 mod 8) = 3 ∧ (-3 mod 8) = 5))
 
Theoremlgsdir2lem2 25830 Lemma for lgsdir2 25834. (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)    &   𝑁𝑆       (𝑁 ∈ ℤ ∧ 2 ∥ (𝑁 + 1) ∧ ((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) → ((𝐴 mod 8) ∈ (0...𝑁) → (𝐴 mod 8) ∈ 𝑆)))
 
Theoremlgsdir2lem3 25831 Lemma for lgsdir2 25834. (Contributed by Mario Carneiro, 4-Feb-2015.)
((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) → (𝐴 mod 8) ∈ ({1, 7} ∪ {3, 5}))
 
Theoremlgsdir2lem4 25832 Lemma for lgsdir2 25834. (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}))
 
Theoremlgsdir2lem5 25833 Lemma for lgsdir2 25834. (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})
 
Theoremlgsdir2 25834 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)))
 
Theoremlgsdirprm 25835 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)))
 
Theoremlgsdir 25836 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). Together with lgsqr 25855 this implies that the product of two quadratic residues or nonresidues is a residue, and the product of a residue and a nonresidue is a nonresidue. (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)))
 
Theoremlgsdilem2 25837* Lemma for lgsdi 25838. (Contributed by Mario Carneiro, 4-Feb-2015.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝑁 ≠ 0)    &   𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, ((𝐴 /L 𝑛)↑(𝑛 pCnt 𝑀)), 1))       (𝜑 → (seq1( · , 𝐹)‘(abs‘𝑀)) = (seq1( · , 𝐹)‘(abs‘(𝑀 · 𝑁))))
 
Theoremlgsdi 25838 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)))
 
Theoremlgsne0 25839 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))
 
Theoremlgsabs1 25840 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))
 
Theoremlgssq 25841 The Legendre symbol at a square is equal to 1. Together with lgsmod 25827 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)
 
Theoremlgssq2 25842 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)
 
Theoremlgsprme0 25843 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))
 
Theorem1lgs 25844 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)
 
Theoremlgs1 25845 The Legendre symbol at 1. See definition in [ApostolNT] p. 188. (Contributed by Mario Carneiro, 28-Apr-2016.)
(𝐴 ∈ ℤ → (𝐴 /L 1) = 1)
 
Theoremlgsmodeq 25846 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)))
 
Theoremlgsmulsqcoprm 25847 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))
 
Theoremlgsdirnn0 25848 Variation on lgsdir 25836 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)))
 
Theoremlgsdinn0 25849 Variation on lgsdi 25838 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 ∈ ℕ0𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐴 /L (𝑀 · 𝑁)) = ((𝐴 /L 𝑀) · (𝐴 /L 𝑁)))
 
Theoremlgsqrlem1 25850 Lemma for lgsqr 25855. (Contributed by Mario Carneiro, 15-Jun-2015.)
𝑌 = (ℤ/nℤ‘𝑃)    &   𝑆 = (Poly1𝑌)    &   𝐵 = (Base‘𝑆)    &   𝐷 = ( deg1𝑌)    &   𝑂 = (eval1𝑌)    &    = (.g‘(mulGrp‘𝑆))    &   𝑋 = (var1𝑌)    &    = (-g𝑆)    &    1 = (1r𝑆)    &   𝑇 = ((((𝑃 − 1) / 2) 𝑋) 1 )    &   𝐿 = (ℤRHom‘𝑌)    &   (𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑 → ((𝐴↑((𝑃 − 1) / 2)) mod 𝑃) = (1 mod 𝑃))       (𝜑 → ((𝑂𝑇)‘(𝐿𝐴)) = (0g𝑌))
 
Theoremlgsqrlem2 25851* Lemma for lgsqr 25855. (Contributed by Mario Carneiro, 15-Jun-2015.)
𝑌 = (ℤ/nℤ‘𝑃)    &   𝑆 = (Poly1𝑌)    &   𝐵 = (Base‘𝑆)    &   𝐷 = ( deg1𝑌)    &   𝑂 = (eval1𝑌)    &    = (.g‘(mulGrp‘𝑆))    &   𝑋 = (var1𝑌)    &    = (-g𝑆)    &    1 = (1r𝑆)    &   𝑇 = ((((𝑃 − 1) / 2) 𝑋) 1 )    &   𝐿 = (ℤRHom‘𝑌)    &   (𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐺 = (𝑦 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘(𝑦↑2)))       (𝜑𝐺:(1...((𝑃 − 1) / 2))–1-1→((𝑂𝑇) “ {(0g𝑌)}))
 
Theoremlgsqrlem3 25852* Lemma for lgsqr 25855. (Contributed by Mario Carneiro, 15-Jun-2015.)
𝑌 = (ℤ/nℤ‘𝑃)    &   𝑆 = (Poly1𝑌)    &   𝐵 = (Base‘𝑆)    &   𝐷 = ( deg1𝑌)    &   𝑂 = (eval1𝑌)    &    = (.g‘(mulGrp‘𝑆))    &   𝑋 = (var1𝑌)    &    = (-g𝑆)    &    1 = (1r𝑆)    &   𝑇 = ((((𝑃 − 1) / 2) 𝑋) 1 )    &   𝐿 = (ℤRHom‘𝑌)    &   (𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐺 = (𝑦 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘(𝑦↑2)))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑 → (𝐴 /L 𝑃) = 1)       (𝜑 → (𝐿𝐴) ∈ ((𝑂𝑇) “ {(0g𝑌)}))
 
Theoremlgsqrlem4 25853* Lemma for lgsqr 25855. (Contributed by Mario Carneiro, 15-Jun-2015.)
𝑌 = (ℤ/nℤ‘𝑃)    &   𝑆 = (Poly1𝑌)    &   𝐵 = (Base‘𝑆)    &   𝐷 = ( deg1𝑌)    &   𝑂 = (eval1𝑌)    &    = (.g‘(mulGrp‘𝑆))    &   𝑋 = (var1𝑌)    &    = (-g𝑆)    &    1 = (1r𝑆)    &   𝑇 = ((((𝑃 − 1) / 2) 𝑋) 1 )    &   𝐿 = (ℤRHom‘𝑌)    &   (𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐺 = (𝑦 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘(𝑦↑2)))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑 → (𝐴 /L 𝑃) = 1)       (𝜑 → ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))
 
Theoremlgsqrlem5 25854* Lemma for lgsqr 25855. (Contributed by Mario Carneiro, 15-Jun-2015.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2}) ∧ (𝐴 /L 𝑃) = 1) → ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))
 
Theoremlgsqr 25855* The Legendre symbol for odd primes is 1 iff the number is not a multiple of the prime (in which case it is 0, see lgsne0 25839) and the number is a quadratic residue mod 𝑃 (it is -1 for nonresidues by the process of elimination from lgsabs1 25840). Given our definition of the Legendre symbol, this theorem is equivalent to Euler's criterion. (Contributed by Mario Carneiro, 15-Jun-2015.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ((𝐴 /L 𝑃) = 1 ↔ (¬ 𝑃𝐴 ∧ ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))))
 
Theoremlgsqrmod 25856* If the Legendre symbol of an integer for an odd prime is 1, then the number is a quadratic residue mod 𝑃. (Contributed by AV, 20-Aug-2021.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ((𝐴 /L 𝑃) = 1 → ∃𝑥 ∈ ℤ ((𝑥↑2) mod 𝑃) = (𝐴 mod 𝑃)))
 
Theoremlgsqrmodndvds 25857* If the Legendre symbol of an integer 𝐴 for an odd prime is 1, then the number is a quadratic residue mod 𝑃 with a solution 𝑥 of the congruence (𝑥↑2)≡𝐴 (mod 𝑃) which is not divisible by the prime. (Contributed by AV, 20-Aug-2021.) (Proof shortened by AV, 18-Mar-2022.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2})) → ((𝐴 /L 𝑃) = 1 → ∃𝑥 ∈ ℤ (((𝑥↑2) mod 𝑃) = (𝐴 mod 𝑃) ∧ ¬ 𝑃𝑥)))
 
Theoremlgsdchrval 25858* The Legendre symbol function 𝑋(𝑚) = (𝑚 /L 𝑁), where 𝑁 is an odd positive number, is a Dirichlet character modulo 𝑁. (Contributed by Mario Carneiro, 28-Apr-2016.)
𝐺 = (DChr‘𝑁)    &   𝑍 = (ℤ/nℤ‘𝑁)    &   𝐷 = (Base‘𝐺)    &   𝐵 = (Base‘𝑍)    &   𝐿 = (ℤRHom‘𝑍)    &   𝑋 = (𝑦𝐵 ↦ (℩𝑚 ∈ ℤ (𝑦 = (𝐿𝑚) ∧ = (𝑚 /L 𝑁))))       (((𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) ∧ 𝐴 ∈ ℤ) → (𝑋‘(𝐿𝐴)) = (𝐴 /L 𝑁))
 
Theoremlgsdchr 25859* The Legendre symbol function 𝑋(𝑚) = (𝑚 /L 𝑁), where 𝑁 is an odd positive number, is a real Dirichlet character modulo 𝑁. (Contributed by Mario Carneiro, 28-Apr-2016.)
𝐺 = (DChr‘𝑁)    &   𝑍 = (ℤ/nℤ‘𝑁)    &   𝐷 = (Base‘𝐺)    &   𝐵 = (Base‘𝑍)    &   𝐿 = (ℤRHom‘𝑍)    &   𝑋 = (𝑦𝐵 ↦ (℩𝑚 ∈ ℤ (𝑦 = (𝐿𝑚) ∧ = (𝑚 /L 𝑁))))       ((𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) → (𝑋𝐷𝑋:𝐵⟶ℝ))
 
14.4.9  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 25878. The general case is still to prove.

 
Theoremgausslemma2dlem0a 25860 Auxiliary lemma 1 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))       (𝜑𝑃 ∈ ℕ)
 
Theoremgausslemma2dlem0b 25861 Auxiliary lemma 2 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑𝐻 ∈ ℕ)
 
Theoremgausslemma2dlem0c 25862 Auxiliary lemma 3 for gausslemma2d 25878. (Contributed by AV, 13-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑 → ((!‘𝐻) gcd 𝑃) = 1)
 
Theoremgausslemma2dlem0d 25863 Auxiliary lemma 4 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))       (𝜑𝑀 ∈ ℕ0)
 
Theoremgausslemma2dlem0e 25864 Auxiliary lemma 5 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))       (𝜑 → (𝑀 · 2) < (𝑃 / 2))
 
Theoremgausslemma2dlem0f 25865 Auxiliary lemma 6 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑 → (𝑀 + 1) ≤ 𝐻)
 
Theoremgausslemma2dlem0g 25866 Auxiliary lemma 7 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑𝑀𝐻)
 
Theoremgausslemma2dlem0h 25867 Auxiliary lemma 8 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑁 = (𝐻𝑀)       (𝜑𝑁 ∈ ℕ0)
 
Theoremgausslemma2dlem0i 25868 Auxiliary lemma 9 for gausslemma2d 25878. (Contributed by AV, 14-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑁 = (𝐻𝑀)       (𝜑 → (((2 /L 𝑃) mod 𝑃) = ((-1↑𝑁) mod 𝑃) → (2 /L 𝑃) = (-1↑𝑁)))
 
Theoremgausslemma2dlem1a 25869* Lemma for gausslemma2dlem1 25870. (Contributed by AV, 1-Jul-2021.)
(𝜑𝑃 ∈ (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))
 
Theoremgausslemma2dlem1 25870* Lemma 1 for gausslemma2d 25878. (Contributed by AV, 5-Jul-2021.)
(𝜑𝑃 ∈ (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)(𝑅𝑘))
 
Theoremgausslemma2dlem2 25871* Lemma 2 for gausslemma2d 25878. (Contributed by AV, 4-Jul-2021.)
(𝜑𝑃 ∈ (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)(𝑅𝑘) = (𝑘 · 2))
 
Theoremgausslemma2dlem3 25872* Lemma 3 for gausslemma2d 25878. (Contributed by AV, 4-Jul-2021.)
(𝜑𝑃 ∈ (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)(𝑅𝑘) = (𝑃 − (𝑘 · 2)))
 
Theoremgausslemma2dlem4 25873* Lemma 4 for gausslemma2d 25878. (Contributed by AV, 16-Jun-2021.)
(𝜑𝑃 ∈ (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)(𝑅𝑘) · ∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅𝑘)))
 
Theoremgausslemma2dlem5a 25874* Lemma for gausslemma2dlem5 25875. (Contributed by AV, 8-Jul-2021.)
(𝜑𝑃 ∈ (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)(𝑅𝑘) mod 𝑃) = (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(-1 · (𝑘 · 2)) mod 𝑃))
 
Theoremgausslemma2dlem5 25875* Lemma 5 for gausslemma2d 25878. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝑁 = (𝐻𝑀)       (𝜑 → (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅𝑘) mod 𝑃) = (((-1↑𝑁) · ∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑘 · 2)) mod 𝑃))
 
Theoremgausslemma2dlem6 25876* Lemma 6 for gausslemma2d 25878. (Contributed by AV, 16-Jun-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝑁 = (𝐻𝑀)       (𝜑 → ((!‘𝐻) mod 𝑃) = ((((-1↑𝑁) · (2↑𝐻)) · (!‘𝐻)) mod 𝑃))
 
Theoremgausslemma2dlem7 25877* Lemma 7 for gausslemma2d 25878. (Contributed by AV, 13-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝑁 = (𝐻𝑀)       (𝜑 → (((-1↑𝑁) · (2↑𝐻)) mod 𝑃) = 1)
 
Theoremgausslemma2d 25878* 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.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝑁 = (𝐻𝑀)       (𝜑 → (2 /L 𝑃) = (-1↑𝑁))
 
14.4.10  Quadratic reciprocity
 
Theoremlgseisenlem1 25879* Lemma for lgseisen 25883. If 𝑅(𝑢) = (𝑄 · 𝑢) mod 𝑃 and 𝑀(𝑢) = (-1↑𝑅(𝑢)) · 𝑅(𝑢), then for any even 1 ≤ 𝑢𝑃 − 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.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))       (𝜑𝑀:(1...((𝑃 − 1) / 2))⟶(1...((𝑃 − 1) / 2)))
 
Theoremlgseisenlem2 25880* Lemma for lgseisen 25883. The function 𝑀 is an injection (and hence a bijection by the pigeonhole principle). (Contributed by Mario Carneiro, 17-Jun-2015.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)       (𝜑𝑀:(1...((𝑃 − 1) / 2))–1-1-onto→(1...((𝑃 − 1) / 2)))
 
Theoremlgseisenlem3 25881* Lemma for lgseisen 25883. (Contributed by Mario Carneiro, 17-Jun-2015.) (Proof shortened by AV, 28-Jul-2019.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   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)))) = (1r𝑌))
 
Theoremlgseisenlem4 25882* Lemma for lgseisen 25883. The function 𝑀 is an injection (and hence a bijection by the pigeonhole principle). (Contributed by Mario Carneiro, 18-Jun-2015.) (Proof shortened by AV, 15-Jun-2019.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   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))
 
Theoremlgseisen 25883* Eisenstein's lemma, an expression for (𝑃 /L 𝑄) when 𝑃, 𝑄 are distinct odd primes. (Contributed by Mario Carneiro, 18-Jun-2015.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)       (𝜑 → (𝑄 /L 𝑃) = (-1↑Σ𝑥 ∈ (1...((𝑃 − 1) / 2))(⌊‘((𝑄 / 𝑃) · (2 · 𝑥)))))
 
Theoremlgsquadlem1 25884* Lemma for lgsquad 25887. Count the members of 𝑆 with odd coordinates. (Contributed by Mario Carneiro, 19-Jun-2015.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   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↑(♯‘{𝑧𝑆 ∣ ¬ 2 ∥ (1st𝑧)})))
 
Theoremlgsquadlem2 25885* Lemma for lgsquad 25887. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 25884 to get the total count of lattice points in 𝑆 (up to parity). (Contributed by Mario Carneiro, 18-Jun-2015.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   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)))
 
Theoremlgsquadlem3 25886* Lemma for lgsquad 25887. (Contributed by Mario Carneiro, 18-Jun-2015.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   (𝜑𝑄 ∈ (ℙ ∖ {2}))    &   (𝜑𝑃𝑄)    &   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)))
 
Theoremlgsquad 25887 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}) ∧ 𝑃𝑄) → ((𝑃 /L 𝑄) · (𝑄 /L 𝑃)) = (-1↑(((𝑃 − 1) / 2) · ((𝑄 − 1) / 2))))
 
Theoremlgsquad2lem1 25888 Lemma for lgsquad2 25890. (Contributed by Mario Carneiro, 19-Jun-2015.)
(𝜑𝑀 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝑀)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝑁)    &   (𝜑 → (𝑀 gcd 𝑁) = 1)    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝐵 ∈ 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))))
 
Theoremlgsquad2lem2 25889* Lemma for lgsquad2 25890. (Contributed by Mario Carneiro, 19-Jun-2015.)
(𝜑𝑀 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝑀)    &   (𝜑𝑁 ∈ 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))))
 
Theoremlgsquad2 25890 Extend lgsquad 25887 to coprime odd integers (the domain of the Jacobi symbol). (Contributed by Mario Carneiro, 19-Jun-2015.)
(𝜑𝑀 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝑀)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝑁)    &   (𝜑 → (𝑀 gcd 𝑁) = 1)       (𝜑 → ((𝑀 /L 𝑁) · (𝑁 /L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))
 
Theoremlgsquad3 25891 Extend lgsquad2 25890 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)))
 
Theoremm1lgs 25892 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))
 
Theorem2lgslem1a1 25893* Lemma 1 for 2lgslem1a 25895. (Contributed by AV, 16-Jun-2021.)
((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))
 
Theorem2lgslem1a2 25894 Lemma 2 for 2lgslem1a 25895. (Contributed by AV, 18-Jun-2021.)
((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))
 
Theorem2lgslem1a 25895* Lemma 1 for 2lgslem1 25898. (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)})
 
Theorem2lgslem1b 25896* Lemma 2 for 2lgslem1 25898. (Contributed by AV, 18-Jun-2021.)
𝐼 = (𝐴...𝐵)    &   𝐹 = (𝑗𝐼 ↦ (𝑗 · 2))       𝐹:𝐼1-1-onto→{𝑥 ∈ ℤ ∣ ∃𝑖𝐼 𝑥 = (𝑖 · 2)}
 
Theorem2lgslem1c 25897 Lemma 3 for 2lgslem1 25898. (Contributed by AV, 19-Jun-2021.)
((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))
 
Theorem2lgslem1 25898* Lemma 1 for 2lgs 25911. (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))))
 
Theorem2lgslem2 25899 Lemma 2 for 2lgs 25911. (Contributed by AV, 20-Jun-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)
 
Theorem2lgslem3a 25900 Lemma for 2lgslem3a1 25904. (Contributed by AV, 14-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝐾 ∈ ℕ0𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))
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