P. 266 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 26501-26600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	lgsdirnn0 26501	Variation on lgsdir 26489 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 26502	Variation on lgsdi 26491 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)))
Theorem	lgsqrlem1 26503	Lemma for lgsqr 26508. (Contributed by Mario Carneiro, 15-Jun-2015.)
⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝑆 = (Poly1‘𝑌)    &   ⊢ 𝐵 = (Base‘𝑆)    &   ⊢ 𝐷 = ( deg1 ‘𝑌)    &   ⊢ 𝑂 = (eval1‘𝑌)    &   ⊢ ↑ = (.g‘(mulGrp‘𝑆))    &   ⊢ 𝑋 = (var1‘𝑌)    &   ⊢ − = (-g‘𝑆)    &   ⊢ 1 = (1r‘𝑆)    &   ⊢ 𝑇 = ((((𝑃 − 1) / 2) ↑ 𝑋) − 1 )    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    &   ⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → ((𝐴↑((𝑃 − 1) / 2)) mod 𝑃) = (1 mod 𝑃))    ⇒   ⊢ (𝜑 → ((𝑂‘𝑇)‘(𝐿‘𝐴)) = (0g‘𝑌))
Theorem	lgsqrlem2 26504*	Lemma for lgsqr 26508. (Contributed by Mario Carneiro, 15-Jun-2015.)
⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝑆 = (Poly1‘𝑌)    &   ⊢ 𝐵 = (Base‘𝑆)    &   ⊢ 𝐷 = ( deg1 ‘𝑌)    &   ⊢ 𝑂 = (eval1‘𝑌)    &   ⊢ ↑ = (.g‘(mulGrp‘𝑆))    &   ⊢ 𝑋 = (var1‘𝑌)    &   ⊢ − = (-g‘𝑆)    &   ⊢ 1 = (1r‘𝑆)    &   ⊢ 𝑇 = ((((𝑃 − 1) / 2) ↑ 𝑋) − 1 )    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    &   ⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐺 = (𝑦 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘(𝑦↑2)))    ⇒   ⊢ (𝜑 → 𝐺:(1...((𝑃 − 1) / 2))–1-1→(◡(𝑂‘𝑇) “ {(0g‘𝑌)}))
Theorem	lgsqrlem3 26505*	Lemma for lgsqr 26508. (Contributed by Mario Carneiro, 15-Jun-2015.)
⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝑆 = (Poly1‘𝑌)    &   ⊢ 𝐵 = (Base‘𝑆)    &   ⊢ 𝐷 = ( deg1 ‘𝑌)    &   ⊢ 𝑂 = (eval1‘𝑌)    &   ⊢ ↑ = (.g‘(mulGrp‘𝑆))    &   ⊢ 𝑋 = (var1‘𝑌)    &   ⊢ − = (-g‘𝑆)    &   ⊢ 1 = (1r‘𝑆)    &   ⊢ 𝑇 = ((((𝑃 − 1) / 2) ↑ 𝑋) − 1 )    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    &   ⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐺 = (𝑦 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘(𝑦↑2)))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → (𝐴 /L 𝑃) = 1)    ⇒   ⊢ (𝜑 → (𝐿‘𝐴) ∈ (◡(𝑂‘𝑇) “ {(0g‘𝑌)}))
Theorem	lgsqrlem4 26506*	Lemma for lgsqr 26508. (Contributed by Mario Carneiro, 15-Jun-2015.)
⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝑆 = (Poly1‘𝑌)    &   ⊢ 𝐵 = (Base‘𝑆)    &   ⊢ 𝐷 = ( deg1 ‘𝑌)    &   ⊢ 𝑂 = (eval1‘𝑌)    &   ⊢ ↑ = (.g‘(mulGrp‘𝑆))    &   ⊢ 𝑋 = (var1‘𝑌)    &   ⊢ − = (-g‘𝑆)    &   ⊢ 1 = (1r‘𝑆)    &   ⊢ 𝑇 = ((((𝑃 − 1) / 2) ↑ 𝑋) − 1 )    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    &   ⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐺 = (𝑦 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘(𝑦↑2)))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → (𝐴 /L 𝑃) = 1)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))
Theorem	lgsqrlem5 26507*	Lemma for lgsqr 26508. (Contributed by Mario Carneiro, 15-Jun-2015.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑃 ∈ (ℙ ∖ {2}) ∧ (𝐴 /L 𝑃) = 1) → ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))
Theorem	lgsqr 26508*	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 26492) and the number is a quadratic residue mod 𝑃 (it is -1 for nonresidues by the process of elimination from lgsabs1 26493). 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 ↔ (¬ 𝑃 ∥ 𝐴 ∧ ∃𝑥 ∈ ℤ 𝑃 ∥ ((𝑥↑2) − 𝐴))))
Theorem	lgsqrmod 26509*	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)))
Theorem	lgsqrmodndvds 26510*	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) ∧ ¬ 𝑃 ∥ 𝑥)))
Theorem	lgsdchrval 26511*	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 = (𝑦 ∈ 𝐵 ↦ (℩ℎ∃𝑚 ∈ ℤ (𝑦 = (𝐿‘𝑚) ∧ ℎ = (𝑚 /L 𝑁))))    ⇒   ⊢ (((𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) ∧ 𝐴 ∈ ℤ) → (𝑋‘(𝐿‘𝐴)) = (𝐴 /L 𝑁))
Theorem	lgsdchr 26512*	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 = (𝑦 ∈ 𝐵 ↦ (℩ℎ∃𝑚 ∈ ℤ (𝑦 = (𝐿‘𝑚) ∧ ℎ = (𝑚 /L 𝑁))))    ⇒   ⊢ ((𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) → (𝑋 ∈ 𝐷 ∧ 𝑋: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 26531. The general case is still to prove.
Theorem	gausslemma2dlem0a 26513	Auxiliary lemma 1 for gausslemma2d 26531. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    ⇒   ⊢ (𝜑 → 𝑃 ∈ ℕ)
Theorem	gausslemma2dlem0b 26514	Auxiliary lemma 2 for gausslemma2d 26531. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝐻 ∈ ℕ)
Theorem	gausslemma2dlem0c 26515	Auxiliary lemma 3 for gausslemma2d 26531. (Contributed by AV, 13-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → ((!‘𝐻) gcd 𝑃) = 1)
Theorem	gausslemma2dlem0d 26516	Auxiliary lemma 4 for gausslemma2d 26531. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → 𝑀 ∈ ℕ0)
Theorem	gausslemma2dlem0e 26517	Auxiliary lemma 5 for gausslemma2d 26531. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (𝑀 · 2) < (𝑃 / 2))
Theorem	gausslemma2dlem0f 26518	Auxiliary lemma 6 for gausslemma2d 26531. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → (𝑀 + 1) ≤ 𝐻)
Theorem	gausslemma2dlem0g 26519	Auxiliary lemma 7 for gausslemma2d 26531. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝑀 ≤ 𝐻)
Theorem	gausslemma2dlem0h 26520	Auxiliary lemma 8 for gausslemma2d 26531. (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 26521	Auxiliary lemma 9 for gausslemma2d 26531. (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 26522*	Lemma for gausslemma2dlem1 26523. (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	gausslemma2dlem1 26523*	Lemma 1 for gausslemma2d 26531. (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 26524*	Lemma 2 for gausslemma2d 26531. (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 26525*	Lemma 3 for gausslemma2d 26531. (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 26526*	Lemma 4 for gausslemma2d 26531. (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 26527*	Lemma for gausslemma2dlem5 26528. (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 26528*	Lemma 5 for gausslemma2d 26531. (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 26529*	Lemma 6 for gausslemma2d 26531. (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 26530*	Lemma 7 for gausslemma2d 26531. (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 26531*	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))
14.4.10  Quadratic reciprocity
Theorem	lgseisenlem1 26532*	Lemma for lgseisen 26536. 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 26533*	Lemma for lgseisen 26536. 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 26534*	Lemma for lgseisen 26536. (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 26535*	Lemma for lgseisen 26536. 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.)
⊢ (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 26536*	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	lgsquadlem1 26537*	Lemma for lgsquad 26540. 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 26538*	Lemma for lgsquad 26540. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 26537 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 26539*	Lemma for lgsquad 26540. (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 26540	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 26541	Lemma for lgsquad2 26543. (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 26542*	Lemma for lgsquad2 26543. (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 26543	Extend lgsquad 26540 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 26544	Extend lgsquad2 26543 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 26545	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 26546*	Lemma 1 for 2lgslem1a 26548. (Contributed by AV, 16-Jun-2021.)
⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))
Theorem	2lgslem1a2 26547	Lemma 2 for 2lgslem1a 26548. (Contributed by AV, 18-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))
Theorem	2lgslem1a 26548*	Lemma 1 for 2lgslem1 26551. (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 26549*	Lemma 2 for 2lgslem1 26551. (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 26550	Lemma 3 for 2lgslem1 26551. (Contributed by AV, 19-Jun-2021.)
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))
Theorem	2lgslem1 26551*	Lemma 1 for 2lgs 26564. (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 26552	Lemma 2 for 2lgs 26564. (Contributed by AV, 20-Jun-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)
Theorem	2lgslem3a 26553	Lemma for 2lgslem3a1 26557. (Contributed by AV, 14-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))
Theorem	2lgslem3b 26554	Lemma for 2lgslem3b1 26558. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3c 26555	Lemma for 2lgslem3c1 26559. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3d 26556	Lemma for 2lgslem3d1 26560. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2))
Theorem	2lgslem3a1 26557	Lemma 1 for 2lgslem3 26561. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0)
Theorem	2lgslem3b1 26558	Lemma 2 for 2lgslem3 26561. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1)
Theorem	2lgslem3c1 26559	Lemma 3 for 2lgslem3 26561. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1)
Theorem	2lgslem3d1 26560	Lemma 4 for 2lgslem3 26561. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0)
Theorem	2lgslem3 26561	Lemma 3 for 2lgs 26564. (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))
Theorem	2lgs2 26562	The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.)
⊢ (2 /L 2) = 0
Theorem	2lgslem4 26563	Lemma 4 for 2lgs 26564: special case of 2lgs 26564 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.)
⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7})
Theorem	2lgs 26564	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 26471) 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}))
Theorem	2lgsoddprmlem1 26565	Lemma 1 for 2lgsoddprm 26573. (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)))
Theorem	2lgsoddprmlem2 26566	Lemma 2 for 2lgsoddprm 26573. (Contributed by AV, 19-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ 2 ∥ (((𝑅↑2) − 1) / 8)))
Theorem	2lgsoddprmlem3a 26567	Lemma 1 for 2lgsoddprmlem3 26571. (Contributed by AV, 20-Jul-2021.)
⊢ (((1↑2) − 1) / 8) = 0
Theorem	2lgsoddprmlem3b 26568	Lemma 2 for 2lgsoddprmlem3 26571. (Contributed by AV, 20-Jul-2021.)
⊢ (((3↑2) − 1) / 8) = 1
Theorem	2lgsoddprmlem3c 26569	Lemma 3 for 2lgsoddprmlem3 26571. (Contributed by AV, 20-Jul-2021.)
⊢ (((5↑2) − 1) / 8) = 3
Theorem	2lgsoddprmlem3d 26570	Lemma 4 for 2lgsoddprmlem3 26571. (Contributed by AV, 20-Jul-2021.)
⊢ (((7↑2) − 1) / 8) = (2 · 3)
Theorem	2lgsoddprmlem3 26571	Lemma 3 for 2lgsoddprm 26573. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7}))
Theorem	2lgsoddprmlem4 26572	Lemma 4 for 2lgsoddprm 26573. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7}))
Theorem	2lgsoddprm 26573	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
Theorem	2sqlem1 26574*	Lemma for 2sq 26587. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑥 ∈ ℤ[i] 𝐴 = ((abs‘𝑥)↑2))
Theorem	2sqlem2 26575*	Lemma for 2sq 26587. (Contributed by Mario Carneiro, 19-Jun-2015.)
⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝐴 = ((𝑥↑2) + (𝑦↑2)))
Theorem	mul2sq 26576	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))    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆) → (𝐴 · 𝐵) ∈ 𝑆)
Theorem	2sqlem3 26577	Lemma for 2sqlem5 26579. (Contributed by Mario Carneiro, 20-Jun-2015.)
⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ (𝜑 → (𝑁 · 𝑃) = ((𝐴↑2) + (𝐵↑2)))    &   ⊢ (𝜑 → 𝑃 = ((𝐶↑2) + (𝐷↑2)))    &   ⊢ (𝜑 → 𝑃 ∥ ((𝐶 · 𝐵) + (𝐴 · 𝐷)))    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝑆)
Theorem	2sqlem4 26578	Lemma for 2sqlem5 26579. (Contributed by Mario Carneiro, 20-Jun-2015.)
⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ (𝜑 → (𝑁 · 𝑃) = ((𝐴↑2) + (𝐵↑2)))    &   ⊢ (𝜑 → 𝑃 = ((𝐶↑2) + (𝐷↑2)))    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝑆)
Theorem	2sqlem5 26579	Lemma for 2sq 26587. 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))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (𝑁 · 𝑃) ∈ 𝑆)    &   ⊢ (𝜑 → 𝑃 ∈ 𝑆)    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝑆)
Theorem	2sqlem6 26580*	Lemma for 2sq 26587. 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))    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → ∀𝑝 ∈ ℙ (𝑝 ∥ 𝐵 → 𝑝 ∈ 𝑆))    &   ⊢ (𝜑 → (𝐴 · 𝐵) ∈ 𝑆)    ⇒   ⊢ (𝜑 → 𝐴 ∈ 𝑆)
Theorem	2sqlem7 26581*	Lemma for 2sq 26587. (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)
Theorem	2sqlem8a 26582*	Lemma for 2sqlem8 26583. (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))∀𝑎 ∈ 𝑌 (𝑏 ∥ 𝑎 → 𝑏 ∈ 𝑆))    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → 𝑁 = ((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)
Theorem	2sqlem8 26583*	Lemma for 2sq 26587. (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))∀𝑎 ∈ 𝑌 (𝑏 ∥ 𝑎 → 𝑏 ∈ 𝑆))    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → 𝑁 = ((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))    ⇒   ⊢ (𝜑 → 𝑀 ∈ 𝑆)
Theorem	2sqlem9 26584*	Lemma for 2sq 26587. (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))∀𝑎 ∈ 𝑌 (𝑏 ∥ 𝑎 → 𝑏 ∈ 𝑆))    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑌)    ⇒   ⊢ (𝜑 → 𝑀 ∈ 𝑆)
Theorem	2sqlem10 26585*	Lemma for 2sq 26587. 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)))}    ⇒   ⊢ ((𝐴 ∈ 𝑌 ∧ 𝐵 ∈ ℕ ∧ 𝐵 ∥ 𝐴) → 𝐵 ∈ 𝑆)
Theorem	2sqlem11 26586*	Lemma for 2sq 26587. (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) → 𝑃 ∈ 𝑆)
Theorem	2sq 26587*	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)))
Theorem	2sqblem 26588	Lemma for 2sqb 26589. (Contributed by Mario Carneiro, 20-Jun-2015.)
⊢ (𝜑 → (𝑃 ∈ ℙ ∧ 𝑃 ≠ 2))    &   ⊢ (𝜑 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))    &   ⊢ (𝜑 → 𝑃 = ((𝑋↑2) + (𝑌↑2)))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → (𝑃 gcd 𝑌) = ((𝑃 · 𝐴) + (𝑌 · 𝐵)))    ⇒   ⊢ (𝜑 → (𝑃 mod 4) = 1)
Theorem	2sqb 26589*	The converse to 2sq 26587. (Contributed by Mario Carneiro, 20-Jun-2015.)
⊢ (𝑃 ∈ ℙ → (∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑃 = ((𝑥↑2) + (𝑦↑2)) ↔ (𝑃 = 2 ∨ (𝑃 mod 4) = 1)))
Theorem	2sq2 26590	2 is the sum of squares of two nonnegative integers iff the two integers are 1. (Contributed by AV, 19-Jun-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → (((𝐴↑2) + (𝐵↑2)) = 2 ↔ (𝐴 = 1 ∧ 𝐵 = 1)))
Theorem	2sqn0 26591	If the sum of two squares is prime, none of the original number is zero. (Contributed by Thierry Arnoux, 4-Feb-2020.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → ((𝐴↑2) + (𝐵↑2)) = 𝑃)    ⇒   ⊢ (𝜑 → 𝐴 ≠ 0)
Theorem	2sqcoprm 26592	If the sum of two squares is prime, the two original numbers are coprime. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → ((𝐴↑2) + (𝐵↑2)) = 𝑃)    ⇒   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)
Theorem	2sqmod 26593	Given two decompositions of a prime as a sum of two squares, show that they are equal. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐴 ≤ 𝐵)    &   ⊢ (𝜑 → 𝐶 ≤ 𝐷)    &   ⊢ (𝜑 → ((𝐴↑2) + (𝐵↑2)) = 𝑃)    &   ⊢ (𝜑 → ((𝐶↑2) + (𝐷↑2)) = 𝑃)    ⇒   ⊢ (𝜑 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
Theorem	2sqmo 26594*	There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 26589 for the existence of such a decomposition. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ (𝑃 ∈ ℙ → ∃*𝑎 ∈ ℕ0 ∃𝑏 ∈ ℕ0 (𝑎 ≤ 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))
Theorem	2sqnn0 26595*	All primes of the form 4𝑘 + 1 are sums of squares of two nonnegative integers. (Contributed by AV, 3-Jun-2023.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃𝑥 ∈ ℕ0 ∃𝑦 ∈ ℕ0 𝑃 = ((𝑥↑2) + (𝑦↑2)))
Theorem	2sqnn 26596*	All primes of the form 4𝑘 + 1 are sums of squares of two positive integers. (Contributed by AV, 11-Jun-2023.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃𝑥 ∈ ℕ ∃𝑦 ∈ ℕ 𝑃 = ((𝑥↑2) + (𝑦↑2)))
Theorem	addsq2reu 26597*	For each complex number 𝐶, there exists a unique complex number 𝑎 added to the square of a unique another complex number 𝑏 resulting in the given complex number 𝐶. The unique complex number 𝑎 is 𝐶, and the unique another complex number 𝑏 is 0. Remark: This, together with addsqnreup 26600, is an example showing that the pattern ∃!𝑎 ∈ 𝐴∃!𝑏 ∈ 𝐵𝜑 does not necessarily mean "There are unique sets 𝑎 and 𝑏 fulfilling 𝜑). See also comments for df-eu 2570 and 2eu4 2657. For more details see comment for addsqnreup 26600. (Contributed by AV, 21-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqn2reu 26598*	For each complex number 𝐶, there does not exist a unique complex number 𝑏, squared and added to a unique another complex number 𝑎 resulting in the given complex number 𝐶. Actually, for each complex number 𝑏, 𝑎 = (𝐶 − (𝑏↑2)) is unique. Remark: This, together with addsq2reu 26597, shows that commutation of two unique quantifications need not be equivalent, and provides an evident justification of the fact that considering the pair of variables is necessary to obtain what we intuitively understand as "double unique existence". (Proposed by GL, 23-Jun-2023.). (Contributed by AV, 23-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑏 ∈ ℂ ∃!𝑎 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqrexnreu 26599*	For each complex number, there exists a complex number to which the square of more than one (or no) other complex numbers can be added to result in the given complex number. Remark: This theorem, together with addsq2reu 26597, shows that there are cases in which there is a set together with a not unique other set fulfilling a wff, although there is a unique set fulfilling the wff together with another unique set (see addsq2reu 26597). For more details see comment for addsqnreup 26600. (Contributed by AV, 20-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqnreup 26600*	There is no unique decomposition of a complex number as a sum of a complex number and a square of a complex number. Remark: This theorem, together with addsq2reu 26597, is a real life example (about a numerical property) showing that the pattern ∃!𝑎 ∈ 𝐴∃!𝑏 ∈ 𝐵𝜑 does not necessarily mean "There are unique sets 𝑎 and 𝑏 fulfilling 𝜑"). See also comments for df-eu 2570 and 2eu4 2657. In the case of decompositions of complex numbers as a sum of a complex number and a square of a complex number, the only/unique complex number to which the square of a unique complex number is added yields in the given complex number is the given number itself, and the unique complex number to be squared is 0 (see comment for addsq2reu 26597). There are, however, complex numbers to which the square of more than one other complex numbers can be added to yield the given complex number (see addsqrexnreu 26599). For example, ⟨1, (√‘(𝐶 − 1))⟩ and ⟨1, -(√‘(𝐶 − 1))⟩ are two different decompositions of 𝐶 (if 𝐶 ≠ 1). Therefore, there is no unique decomposition of any complex number as a sum of a complex number and a square of a complex number, as generally proved by this theorem. As a consequence, a theorem must claim the existence of a unique pair of sets to express "There are unique 𝑎 and 𝑏 so that .." (more formally ∃!𝑝 ∈ (𝐴 × 𝐵)𝜑 with 𝑝 = ⟨𝑎, 𝑏⟩), or by showing (∃!𝑥 ∈ 𝐴∃𝑦 ∈ 𝐵𝜑 ∧ ∃!𝑦 ∈ 𝐵∃𝑥 ∈ 𝐴𝜑) (see 2reu4 4458 resp. 2eu4 2657). These two representations are equivalent (see opreu2reurex 6201). An analogon of this theorem using the latter variant is given in addsqn2reurex2 26602. In some cases, however, the variant with (ordered!) pairs may be possible only for ordered sets (like ℝ or ℙ) and claiming that the first component is less than or equal to the second component (see, for example, 2sqreunnltb 26618 and 2sqreuopb 26625). Alternatively, (proper) unordered pairs can be used: ∃!𝑝𝑒𝒫 𝐴((♯‘𝑝) = 2 ∧ 𝜑), or, using the definition of proper pairs: ∃!𝑝 ∈ (Pairsproper‘𝐴)𝜑 (see, for example, inlinecirc02preu 46145). (Contributed by AV, 21-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶)