P. 273 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 27201-27300   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	lgsqrmod 27201*	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 27202*	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 27203*	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 27204*	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 27223. The general case is still to prove.
Theorem	gausslemma2dlem0a 27205	Auxiliary lemma 1 for gausslemma2d 27223. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    ⇒   ⊢ (𝜑 → 𝑃 ∈ ℕ)
Theorem	gausslemma2dlem0b 27206	Auxiliary lemma 2 for gausslemma2d 27223. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝐻 ∈ ℕ)
Theorem	gausslemma2dlem0c 27207	Auxiliary lemma 3 for gausslemma2d 27223. (Contributed by AV, 13-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → ((!‘𝐻) gcd 𝑃) = 1)
Theorem	gausslemma2dlem0d 27208	Auxiliary lemma 4 for gausslemma2d 27223. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → 𝑀 ∈ ℕ0)
Theorem	gausslemma2dlem0e 27209	Auxiliary lemma 5 for gausslemma2d 27223. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (𝑀 · 2) < (𝑃 / 2))
Theorem	gausslemma2dlem0f 27210	Auxiliary lemma 6 for gausslemma2d 27223. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → (𝑀 + 1) ≤ 𝐻)
Theorem	gausslemma2dlem0g 27211	Auxiliary lemma 7 for gausslemma2d 27223. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝑀 ≤ 𝐻)
Theorem	gausslemma2dlem0h 27212	Auxiliary lemma 8 for gausslemma2d 27223. (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 27213	Auxiliary lemma 9 for gausslemma2d 27223. (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 27214*	Lemma for gausslemma2dlem1 27215. (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 27215*	Lemma 1 for gausslemma2d 27223. (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 27216*	Lemma 2 for gausslemma2d 27223. (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 27217*	Lemma 3 for gausslemma2d 27223. (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 27218*	Lemma 4 for gausslemma2d 27223. (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 27219*	Lemma for gausslemma2dlem5 27220. (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 27220*	Lemma 5 for gausslemma2d 27223. (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 27221*	Lemma 6 for gausslemma2d 27223. (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 27222*	Lemma 7 for gausslemma2d 27223. (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 27223*	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 27224*	Lemma for lgseisen 27228. 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 27225*	Lemma for lgseisen 27228. 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 27226*	Lemma for lgseisen 27228. (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 27227*	Lemma for lgseisen 27228. 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 27228*	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 27229*	Lemma for lgsquad 27232. 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 27230*	Lemma for lgsquad 27232. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27229 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 27231*	Lemma for lgsquad 27232. (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 27232	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 27233	Lemma for lgsquad2 27235. (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 27234*	Lemma for lgsquad2 27235. (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 27235	Extend lgsquad 27232 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 27236	Extend lgsquad2 27235 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 27237	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 27238*	Lemma 1 for 2lgslem1a 27240. (Contributed by AV, 16-Jun-2021.)
⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))
Theorem	2lgslem1a2 27239	Lemma 2 for 2lgslem1a 27240. (Contributed by AV, 18-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))
Theorem	2lgslem1a 27240*	Lemma 1 for 2lgslem1 27243. (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 27241*	Lemma 2 for 2lgslem1 27243. (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 27242	Lemma 3 for 2lgslem1 27243. (Contributed by AV, 19-Jun-2021.)
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))
Theorem	2lgslem1 27243*	Lemma 1 for 2lgs 27256. (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 27244	Lemma 2 for 2lgs 27256. (Contributed by AV, 20-Jun-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)
Theorem	2lgslem3a 27245	Lemma for 2lgslem3a1 27249. (Contributed by AV, 14-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))
Theorem	2lgslem3b 27246	Lemma for 2lgslem3b1 27250. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3c 27247	Lemma for 2lgslem3c1 27251. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3d 27248	Lemma for 2lgslem3d1 27252. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2))
Theorem	2lgslem3a1 27249	Lemma 1 for 2lgslem3 27253. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0)
Theorem	2lgslem3b1 27250	Lemma 2 for 2lgslem3 27253. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1)
Theorem	2lgslem3c1 27251	Lemma 3 for 2lgslem3 27253. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1)
Theorem	2lgslem3d1 27252	Lemma 4 for 2lgslem3 27253. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0)
Theorem	2lgslem3 27253	Lemma 3 for 2lgs 27256. (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 27254	The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.)
⊢ (2 /L 2) = 0
Theorem	2lgslem4 27255	Lemma 4 for 2lgs 27256: special case of 2lgs 27256 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.)
⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7})
Theorem	2lgs 27256	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 27163) 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 27257	Lemma 1 for 2lgsoddprm 27265. (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 27258	Lemma 2 for 2lgsoddprm 27265. (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 27259	Lemma 1 for 2lgsoddprmlem3 27263. (Contributed by AV, 20-Jul-2021.)
⊢ (((1↑2) − 1) / 8) = 0
Theorem	2lgsoddprmlem3b 27260	Lemma 2 for 2lgsoddprmlem3 27263. (Contributed by AV, 20-Jul-2021.)
⊢ (((3↑2) − 1) / 8) = 1
Theorem	2lgsoddprmlem3c 27261	Lemma 3 for 2lgsoddprmlem3 27263. (Contributed by AV, 20-Jul-2021.)
⊢ (((5↑2) − 1) / 8) = 3
Theorem	2lgsoddprmlem3d 27262	Lemma 4 for 2lgsoddprmlem3 27263. (Contributed by AV, 20-Jul-2021.)
⊢ (((7↑2) − 1) / 8) = (2 · 3)
Theorem	2lgsoddprmlem3 27263	Lemma 3 for 2lgsoddprm 27265. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7}))
Theorem	2lgsoddprmlem4 27264	Lemma 4 for 2lgsoddprm 27265. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7}))
Theorem	2lgsoddprm 27265	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 27266*	Lemma for 2sq 27279. (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 27267*	Lemma for 2sq 27279. (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 27268	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 27269	Lemma for 2sqlem5 27271. (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 27270	Lemma for 2sqlem5 27271. (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 27271	Lemma for 2sq 27279. 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 27272*	Lemma for 2sq 27279. 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 27273*	Lemma for 2sq 27279. (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 27274*	Lemma for 2sqlem8 27275. (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 27275*	Lemma for 2sq 27279. (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 27276*	Lemma for 2sq 27279. (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 27277*	Lemma for 2sq 27279. 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 27278*	Lemma for 2sq 27279. (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 27279*	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 27280	Lemma for 2sqb 27281. (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 27281*	The converse to 2sq 27279. (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 27282	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 27283	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 27284	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 27285	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 27286*	There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27281 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 27287*	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 27288*	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 27289*	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 27292, 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 2555 and 2eu4 2642. For more details see comment for addsqnreup 27292. (Contributed by AV, 21-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqn2reu 27290*	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 27289, 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 27291*	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 27289, 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 27289). For more details see comment for addsqnreup 27292. (Contributed by AV, 20-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqnreup 27292*	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 27289, 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 2555 and 2eu4 2642. 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 27289). 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 27291). 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 4518 resp. 2eu4 2642). These two representations are equivalent (see opreu2reurex 6283). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27294. 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 27310 and 2sqreuopb 27317). 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 47662). (Contributed by AV, 21-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶)
Theorem	addsq2nreurex 27293*	For each complex number 𝐶, there is no unique complex number 𝑎 added to the square of another complex number 𝑏 resulting in the given complex number 𝐶. (Contributed by AV, 2-Jul-2023.)
⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑎 ∈ ℂ ∃𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqn2reurex2 27294*	For each complex number 𝐶, there does not uniquely exist two complex numbers 𝑎 and 𝑏, with 𝑏 squared and added to 𝑎 resulting in the given complex number 𝐶. Remark: This, together with addsq2reu 27289, is an example showing that the pattern ∃!𝑎 ∈ 𝐴∃!𝑏 ∈ 𝐵𝜑 does not necessarily mean "There are unique sets 𝑎 and 𝑏 fulfilling 𝜑), as it is the case with the pattern (∃!𝑎 ∈ 𝐴∃𝑏 ∈ 𝐵𝜑 ∧ ∃!𝑏 ∈ 𝐵∃𝑎 ∈ 𝐴𝜑. See also comments for df-eu 2555 and 2eu4 2642. (Contributed by AV, 2-Jul-2023.)
⊢ (𝐶 ∈ ℂ → ¬ (∃!𝑎 ∈ ℂ ∃𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶 ∧ ∃!𝑏 ∈ ℂ ∃𝑎 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶))
Theorem	2sqreulem1 27295*	Lemma 1 for 2sqreu 27305. (Contributed by AV, 4-Jun-2023.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑎 ∈ ℕ0 ∃!𝑏 ∈ ℕ0 (𝑎 ≤ 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))
Theorem	2sqreultlem 27296*	Lemma for 2sqreult 27307. (Contributed by AV, 8-Jun-2023.) (Proposed by GL, 8-Jun-2023.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑎 ∈ ℕ0 ∃!𝑏 ∈ ℕ0 (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))
Theorem	2sqreultblem 27297*	Lemma for 2sqreultb 27308. (Contributed by AV, 10-Jun-2023.) The prime needs not be odd, as observed by WL. (Revised by AV, 18-Jun-2023.)
⊢ (𝑃 ∈ ℙ → ((𝑃 mod 4) = 1 ↔ ∃!𝑎 ∈ ℕ0 ∃!𝑏 ∈ ℕ0 (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)))
Theorem	2sqreunnlem1 27298*	Lemma 1 for 2sqreunn 27306. (Contributed by AV, 11-Jun-2023.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑎 ∈ ℕ ∃!𝑏 ∈ ℕ (𝑎 ≤ 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))
Theorem	2sqreunnltlem 27299*	Lemma for 2sqreunnlt 27309. (Contributed by AV, 4-Jun-2023.) Specialization to different integers, proposed by GL. (Revised by AV, 11-Jun-2023.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑎 ∈ ℕ ∃!𝑏 ∈ ℕ (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))
Theorem	2sqreunnltblem 27300*	Lemma for 2sqreunnltb 27310. (Contributed by AV, 11-Jun-2023.) The prime needs not be odd, as observed by WL. (Revised by AV, 18-Jun-2023.)
⊢ (𝑃 ∈ ℙ → ((𝑃 mod 4) = 1 ↔ ∃!𝑎 ∈ ℕ ∃!𝑏 ∈ ℕ (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)))