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Theorem List for Metamath Proof Explorer - 25101-25200   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremlgsdirprm 25101 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 25102 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 25121 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 25103* Lemma for lgsdi 25104. (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 25104 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 25105 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 25106 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 25107 The Legendre symbol at a square is equal to 1. Together with lgsmod 25093 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 25108 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 25109 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 25110 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 25111 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 25112 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 25113 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 25114 Variation on lgsdir 25102 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 25115 Variation on lgsdi 25104 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 25116 Lemma for lgsqr 25121. (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 25117* Lemma for lgsqr 25121. (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 25118* Lemma for lgsqr 25121. (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 25119* Lemma for lgsqr 25121. (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 25120* Lemma for lgsqr 25121. (Contributed by Mario Carneiro, 15-Jun-2015.)
((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2}) ∧ (𝐴 /L 𝑃) = 1) → ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))
 
Theoremlgsqr 25121* 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 25105) and the number is a quadratic residue mod 𝑃 (it is -1 for nonresidues by the process of elimination from lgsabs1 25106). 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 25122* 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 25123* 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 25124* 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 25125* 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 25144. The general case is still to prove.

 
Theoremgausslemma2dlem0a 25126 Auxiliary lemma 1 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))       (𝜑𝑃 ∈ ℕ)
 
Theoremgausslemma2dlem0b 25127 Auxiliary lemma 2 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑𝐻 ∈ ℕ)
 
Theoremgausslemma2dlem0c 25128 Auxiliary lemma 3 for gausslemma2d 25144. (Contributed by AV, 13-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑 → ((!‘𝐻) gcd 𝑃) = 1)
 
Theoremgausslemma2dlem0d 25129 Auxiliary lemma 4 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))       (𝜑𝑀 ∈ ℕ0)
 
Theoremgausslemma2dlem0e 25130 Auxiliary lemma 5 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))       (𝜑 → (𝑀 · 2) < (𝑃 / 2))
 
Theoremgausslemma2dlem0f 25131 Auxiliary lemma 6 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑 → (𝑀 + 1) ≤ 𝐻)
 
Theoremgausslemma2dlem0g 25132 Auxiliary lemma 7 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)       (𝜑𝑀𝐻)
 
Theoremgausslemma2dlem0h 25133 Auxiliary lemma 8 for gausslemma2d 25144. (Contributed by AV, 9-Jul-2021.)
(𝜑𝑃 ∈ (ℙ ∖ {2}))    &   𝑀 = (⌊‘(𝑃 / 4))    &   𝐻 = ((𝑃 − 1) / 2)    &   𝑁 = (𝐻𝑀)       (𝜑𝑁 ∈ ℕ0)
 
Theoremgausslemma2dlem0i 25134 Auxiliary lemma 9 for gausslemma2d 25144. (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 25135* Lemma for gausslemma2dlem1 25136. (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 25136* Lemma 1 for gausslemma2d 25144. (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 25137* Lemma 2 for gausslemma2d 25144. (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 25138* Lemma 3 for gausslemma2d 25144. (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 25139* Lemma 4 for gausslemma2d 25144. (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 25140* Lemma for gausslemma2dlem5 25141. (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 25141* Lemma 5 for gausslemma2d 25144. (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 25142* Lemma 6 for gausslemma2d 25144. (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 25143* Lemma 7 for gausslemma2d 25144. (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 25144* 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 25145* Lemma for lgseisen 25149. 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 25146* Lemma for lgseisen 25149. 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 25147* Lemma for lgseisen 25149. (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 25148* Lemma for lgseisen 25149. 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 25149* 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 25150* Lemma for lgsquad 25153. 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 25151* Lemma for lgsquad 25153. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 25150 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 25152* Lemma for lgsquad 25153. (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 25153 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 25154 Lemma for lgsquad2 25156. (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 25155* Lemma for lgsquad2 25156. (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 25156 Extend lgsquad 25153 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 25157 Extend lgsquad2 25156 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 25158 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 25159* Lemma 1 for 2lgslem1a 25161. (Contributed by AV, 16-Jun-2021.)
((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))
 
Theorem2lgslem1a2 25160 Lemma 2 for 2lgslem1a 25161. (Contributed by AV, 18-Jun-2021.)
((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))
 
Theorem2lgslem1a 25161* Lemma 1 for 2lgslem1 25164. (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 25162* Lemma 2 for 2lgslem1 25164. (Contributed by AV, 18-Jun-2021.)
𝐼 = (𝐴...𝐵)    &   𝐹 = (𝑗𝐼 ↦ (𝑗 · 2))       𝐹:𝐼1-1-onto→{𝑥 ∈ ℤ ∣ ∃𝑖𝐼 𝑥 = (𝑖 · 2)}
 
Theorem2lgslem1c 25163 Lemma 3 for 2lgslem1 25164. (Contributed by AV, 19-Jun-2021.)
((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))
 
Theorem2lgslem1 25164* Lemma 1 for 2lgs 25177. (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 25165 Lemma 2 for 2lgs 25177. (Contributed by AV, 20-Jun-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)
 
Theorem2lgslem3a 25166 Lemma for 2lgslem3a1 25170. (Contributed by AV, 14-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝐾 ∈ ℕ0𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))
 
Theorem2lgslem3b 25167 Lemma for 2lgslem3b1 25171. (Contributed by AV, 16-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝐾 ∈ ℕ0𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1))
 
Theorem2lgslem3c 25168 Lemma for 2lgslem3c1 25172. (Contributed by AV, 16-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝐾 ∈ ℕ0𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1))
 
Theorem2lgslem3d 25169 Lemma for 2lgslem3d1 25173. (Contributed by AV, 16-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝐾 ∈ ℕ0𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2))
 
Theorem2lgslem3a1 25170 Lemma 1 for 2lgslem3 25174. (Contributed by AV, 15-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0)
 
Theorem2lgslem3b1 25171 Lemma 2 for 2lgslem3 25174. (Contributed by AV, 16-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1)
 
Theorem2lgslem3c1 25172 Lemma 3 for 2lgslem3 25174. (Contributed by AV, 16-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1)
 
Theorem2lgslem3d1 25173 Lemma 4 for 2lgslem3 25174. (Contributed by AV, 15-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0)
 
Theorem2lgslem3 25174 Lemma 3 for 2lgs 25177. (Contributed by AV, 16-Jul-2021.)
𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))       ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → (𝑁 mod 2) = if((𝑃 mod 8) ∈ {1, 7}, 0, 1))
 
Theorem2lgs2 25175 The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.)
(2 /L 2) = 0
 
Theorem2lgslem4 25176 Lemma 4 for 2lgs 25177: special case of 2lgs 25177 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.)
((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7})
 
Theorem2lgs 25177 The second supplement to the law of quadratic reciprocity (for the Legendre symbol extended to arbitrary primes as second argument). Two is a square modulo a prime 𝑃 iff 𝑃≡±1 (mod 8), see first case of theorem 9.5 in [ApostolNT] p. 181. This theorem justifies our definition of (𝑁 /L 2) (lgs2 25084) to some degree, by demanding that reciprocity extend to the case 𝑄 = 2. (Proposed by Mario Carneiro, 19-Jun-2015.) (Contributed by AV, 16-Jul-2021.)
(𝑃 ∈ ℙ → ((2 /L 𝑃) = 1 ↔ (𝑃 mod 8) ∈ {1, 7}))
 
Theorem2lgsoddprmlem1 25178 Lemma 1 for 2lgsoddprm 25186. (Contributed by AV, 19-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 = ((8 · 𝐴) + 𝐵)) → (((𝑁↑2) − 1) / 8) = (((8 · (𝐴↑2)) + (2 · (𝐴 · 𝐵))) + (((𝐵↑2) − 1) / 8)))
 
Theorem2lgsoddprmlem2 25179 Lemma 2 for 2lgsoddprm 25186. (Contributed by AV, 19-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ 2 ∥ (((𝑅↑2) − 1) / 8)))
 
Theorem2lgsoddprmlem3a 25180 Lemma 1 for 2lgsoddprmlem3 25184. (Contributed by AV, 20-Jul-2021.)
(((1↑2) − 1) / 8) = 0
 
Theorem2lgsoddprmlem3b 25181 Lemma 2 for 2lgsoddprmlem3 25184. (Contributed by AV, 20-Jul-2021.)
(((3↑2) − 1) / 8) = 1
 
Theorem2lgsoddprmlem3c 25182 Lemma 3 for 2lgsoddprmlem3 25184. (Contributed by AV, 20-Jul-2021.)
(((5↑2) − 1) / 8) = 3
 
Theorem2lgsoddprmlem3d 25183 Lemma 4 for 2lgsoddprmlem3 25184. (Contributed by AV, 20-Jul-2021.)
(((7↑2) − 1) / 8) = (2 · 3)
 
Theorem2lgsoddprmlem3 25184 Lemma 3 for 2lgsoddprm 25186. (Contributed by AV, 20-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7}))
 
Theorem2lgsoddprmlem4 25185 Lemma 4 for 2lgsoddprm 25186. (Contributed by AV, 20-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7}))
 
Theorem2lgsoddprm 25186 The second supplement to the law of quadratic reciprocity for odd primes (common representation, see theorem 9.5 in [ApostolNT] p. 181): The Legendre symbol for 2 at an odd prime is minus one to the power of the square of the odd prime minus one divided by eight ((2 /L 𝑃) = -1^(((P^2)-1)/8) ). (Contributed by AV, 20-Jul-2021.)
(𝑃 ∈ (ℙ ∖ {2}) → (2 /L 𝑃) = (-1↑(((𝑃↑2) − 1) / 8)))
 
14.4.11  All primes 4n+1 are the sum of two squares
 
Theorem2sqlem1 25187* Lemma for 2sq 25200. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))       (𝐴𝑆 ↔ ∃𝑥 ∈ ℤ[i] 𝐴 = ((abs‘𝑥)↑2))
 
Theorem2sqlem2 25188* Lemma for 2sq 25200. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))       (𝐴𝑆 ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝐴 = ((𝑥↑2) + (𝑦↑2)))
 
Theoremmul2sq 25189 Fibonacci's identity (actually due to Diophantus). The product of two sums of two squares is also a sum of two squares. We can take advantage of Gaussian integers here to trivialize the proof. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))       ((𝐴𝑆𝐵𝑆) → (𝐴 · 𝐵) ∈ 𝑆)
 
Theorem2sqlem3 25190 Lemma for 2sqlem5 25192. (Contributed by Mario Carneiro, 20-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   (𝜑 → (𝑁 · 𝑃) = ((𝐴↑2) + (𝐵↑2)))    &   (𝜑𝑃 = ((𝐶↑2) + (𝐷↑2)))    &   (𝜑𝑃 ∥ ((𝐶 · 𝐵) + (𝐴 · 𝐷)))       (𝜑𝑁𝑆)
 
Theorem2sqlem4 25191 Lemma for 2sqlem5 25192. (Contributed by Mario Carneiro, 20-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐶 ∈ ℤ)    &   (𝜑𝐷 ∈ ℤ)    &   (𝜑 → (𝑁 · 𝑃) = ((𝐴↑2) + (𝐵↑2)))    &   (𝜑𝑃 = ((𝐶↑2) + (𝐷↑2)))       (𝜑𝑁𝑆)
 
Theorem2sqlem5 25192 Lemma for 2sq 25200. If a number that is a sum of two squares is divisible by a prime that is a sum of two squares, then the quotient is a sum of two squares. (Contributed by Mario Carneiro, 20-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (𝑁 · 𝑃) ∈ 𝑆)    &   (𝜑𝑃𝑆)       (𝜑𝑁𝑆)
 
Theorem2sqlem6 25193* Lemma for 2sq 25200. If a number that is a sum of two squares is divisible by a number whose prime divisors are all sums of two squares, then the quotient is a sum of two squares. (Contributed by Mario Carneiro, 20-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑 → ∀𝑝 ∈ ℙ (𝑝𝐵𝑝𝑆))    &   (𝜑 → (𝐴 · 𝐵) ∈ 𝑆)       (𝜑𝐴𝑆)
 
Theorem2sqlem7 25194* Lemma for 2sq 25200. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}       𝑌 ⊆ (𝑆 ∩ ℕ)
 
Theorem2sqlem8a 25195* Lemma for 2sqlem8 25196. (Contributed by Mario Carneiro, 4-Jun-2016.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    &   (𝜑 → ∀𝑏 ∈ (1...(𝑀 − 1))∀𝑎𝑌 (𝑏𝑎𝑏𝑆))    &   (𝜑𝑀𝑁)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑𝑁 = ((𝐴↑2) + (𝐵↑2)))    &   𝐶 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐷 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))       (𝜑 → (𝐶 gcd 𝐷) ∈ ℕ)
 
Theorem2sqlem8 25196* Lemma for 2sq 25200. (Contributed by Mario Carneiro, 20-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    &   (𝜑 → ∀𝑏 ∈ (1...(𝑀 − 1))∀𝑎𝑌 (𝑏𝑎𝑏𝑆))    &   (𝜑𝑀𝑁)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ (ℤ‘2))    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑𝑁 = ((𝐴↑2) + (𝐵↑2)))    &   𝐶 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐷 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   𝐸 = (𝐶 / (𝐶 gcd 𝐷))    &   𝐹 = (𝐷 / (𝐶 gcd 𝐷))       (𝜑𝑀𝑆)
 
Theorem2sqlem9 25197* Lemma for 2sq 25200. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    &   (𝜑 → ∀𝑏 ∈ (1...(𝑀 − 1))∀𝑎𝑌 (𝑏𝑎𝑏𝑆))    &   (𝜑𝑀𝑁)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁𝑌)       (𝜑𝑀𝑆)
 
Theorem2sqlem10 25198* Lemma for 2sq 25200. Every factor of a "proper" sum of two squares (where the summands are coprime) is a sum of two squares. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}       ((𝐴𝑌𝐵 ∈ ℕ ∧ 𝐵𝐴) → 𝐵𝑆)
 
Theorem2sqlem11 25199* Lemma for 2sq 25200. (Contributed by Mario Carneiro, 19-Jun-2015.)
𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}       ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → 𝑃𝑆)
 
Theorem2sq 25200* All primes of the form 4𝑘 + 1 are sums of two squares. This is Metamath 100 proof #20. (Contributed by Mario Carneiro, 20-Jun-2015.)
((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑃 = ((𝑥↑2) + (𝑦↑2)))
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