P. 277 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 27601-27700   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	gausslemma2dlem0b 27601	Auxiliary lemma 2 for gausslemma2d 27618. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝐻 ∈ ℕ)
Theorem	gausslemma2dlem0c 27602	Auxiliary lemma 3 for gausslemma2d 27618. (Contributed by AV, 13-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → ((!‘𝐻) gcd 𝑃) = 1)
Theorem	gausslemma2dlem0d 27603	Auxiliary lemma 4 for gausslemma2d 27618. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → 𝑀 ∈ ℕ0)
Theorem	gausslemma2dlem0e 27604	Auxiliary lemma 5 for gausslemma2d 27618. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (𝑀 · 2) < (𝑃 / 2))
Theorem	gausslemma2dlem0f 27605	Auxiliary lemma 6 for gausslemma2d 27618. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → (𝑀 + 1) ≤ 𝐻)
Theorem	gausslemma2dlem0g 27606	Auxiliary lemma 7 for gausslemma2d 27618. (Contributed by AV, 9-Jul-2021.)
⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝑀 ≤ 𝐻)
Theorem	gausslemma2dlem0h 27607	Auxiliary lemma 8 for gausslemma2d 27618. (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 27608	Auxiliary lemma 9 for gausslemma2d 27618. (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 27609*	Lemma for gausslemma2dlem1 27610. (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 27610*	Lemma 1 for gausslemma2d 27618. (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 27611*	Lemma 2 for gausslemma2d 27618. (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 27612*	Lemma 3 for gausslemma2d 27618. (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 27613*	Lemma 4 for gausslemma2d 27618. (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 27614*	Lemma for gausslemma2dlem5 27615. (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 27615*	Lemma 5 for gausslemma2d 27618. (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 27616*	Lemma 6 for gausslemma2d 27618. (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 27617*	Lemma 7 for gausslemma2d 27618. (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 27618*	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 27619*	Lemma for lgseisen 27623. 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 27620*	Lemma for lgseisen 27623. 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 27621*	Lemma for lgseisen 27623. (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 27622*	Lemma for lgseisen 27623. (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 27623*	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 27624*	Lemma for lgsquad 27627. 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 27625*	Lemma for lgsquad 27627. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27624 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 27626*	Lemma for lgsquad 27627. (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 27627	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 27628	Lemma for lgsquad2 27630. (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 27629*	Lemma for lgsquad2 27630. (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 27630	Extend lgsquad 27627 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 27631	Extend lgsquad2 27630 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 27632	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 27633*	Lemma 1 for 2lgslem1a 27635. (Contributed by AV, 16-Jun-2021.)
⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))
Theorem	2lgslem1a2 27634	Lemma 2 for 2lgslem1a 27635. (Contributed by AV, 18-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))
Theorem	2lgslem1a 27635*	Lemma 1 for 2lgslem1 27638. (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 27636*	Lemma 2 for 2lgslem1 27638. (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 27637	Lemma 3 for 2lgslem1 27638. (Contributed by AV, 19-Jun-2021.)
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))
Theorem	2lgslem1 27638*	Lemma 1 for 2lgs 27651. (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 27639	Lemma 2 for 2lgs 27651. (Contributed by AV, 20-Jun-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)
Theorem	2lgslem3a 27640	Lemma for 2lgslem3a1 27644. (Contributed by AV, 14-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))
Theorem	2lgslem3b 27641	Lemma for 2lgslem3b1 27645. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3c 27642	Lemma for 2lgslem3c1 27646. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1))
Theorem	2lgslem3d 27643	Lemma for 2lgslem3d1 27647. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2))
Theorem	2lgslem3a1 27644	Lemma 1 for 2lgslem3 27648. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0)
Theorem	2lgslem3b1 27645	Lemma 2 for 2lgslem3 27648. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1)
Theorem	2lgslem3c1 27646	Lemma 3 for 2lgslem3 27648. (Contributed by AV, 16-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1)
Theorem	2lgslem3d1 27647	Lemma 4 for 2lgslem3 27648. (Contributed by AV, 15-Jul-2021.)
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0)
Theorem	2lgslem3 27648	Lemma 3 for 2lgs 27651. (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 27649	The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.)
⊢ (2 /L 2) = 0
Theorem	2lgslem4 27650	Lemma 4 for 2lgs 27651: special case of 2lgs 27651 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.)
⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7})
Theorem	2lgs 27651	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 27558) 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 27652	Lemma 1 for 2lgsoddprm 27660. (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 27653	Lemma 2 for 2lgsoddprm 27660. (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 27654	Lemma 1 for 2lgsoddprmlem3 27658. (Contributed by AV, 20-Jul-2021.)
⊢ (((1↑2) − 1) / 8) = 0
Theorem	2lgsoddprmlem3b 27655	Lemma 2 for 2lgsoddprmlem3 27658. (Contributed by AV, 20-Jul-2021.)
⊢ (((3↑2) − 1) / 8) = 1
Theorem	2lgsoddprmlem3c 27656	Lemma 3 for 2lgsoddprmlem3 27658. (Contributed by AV, 20-Jul-2021.)
⊢ (((5↑2) − 1) / 8) = 3
Theorem	2lgsoddprmlem3d 27657	Lemma 4 for 2lgsoddprmlem3 27658. (Contributed by AV, 20-Jul-2021.)
⊢ (((7↑2) − 1) / 8) = (2 · 3)
Theorem	2lgsoddprmlem3 27658	Lemma 3 for 2lgsoddprm 27660. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7}))
Theorem	2lgsoddprmlem4 27659	Lemma 4 for 2lgsoddprm 27660. (Contributed by AV, 20-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7}))
Theorem	2lgsoddprm 27660	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 27661*	Lemma for 2sq 27674. (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 27662*	Lemma for 2sq 27674. (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 27663	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 27664	Lemma for 2sqlem5 27666. (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 27665	Lemma for 2sqlem5 27666. (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 27666	Lemma for 2sq 27674. 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 27667*	Lemma for 2sq 27674. 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 27668*	Lemma for 2sq 27674. (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 27669*	Lemma for 2sqlem8 27670. (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 27670*	Lemma for 2sq 27674. (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 27671*	Lemma for 2sq 27674. (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 27672*	Lemma for 2sq 27674. 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 27673*	Lemma for 2sq 27674. (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 27674*	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 27675	Lemma for 2sqb 27676. (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 27676*	The converse to 2sq 27674. (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 27677	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 27678	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 27679	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 27680	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 27681*	There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27676 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 27682*	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 27683*	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 27684*	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 27687, 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 2596 and 2eu4 2681. For more details see comment for addsqnreup 27687. (Contributed by AV, 21-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqn2reu 27685*	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 27684, 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 27686*	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 27684, 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 27684). For more details see comment for addsqnreup 27687. (Contributed by AV, 20-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶)
Theorem	addsqnreup 27687*	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 27684, 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 2596 and 2eu4 2681. 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 27684). 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 27686). 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 4483 resp. 2eu4 2681). These two representations are equivalent (see opreu2reurex 6296). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27689. 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 27705 and 2sqreuopb 27712). 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 49726). (Contributed by AV, 21-Jun-2023.)
⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶)
Theorem	addsq2nreurex 27688*	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 27689*	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 27684, 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 2596 and 2eu4 2681. (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 27690*	Lemma 1 for 2sqreu 27700. (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 27691*	Lemma for 2sqreult 27702. (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 27692*	Lemma for 2sqreultb 27703. (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 27693*	Lemma 1 for 2sqreunn 27701. (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 27694*	Lemma for 2sqreunnlt 27704. (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 27695*	Lemma for 2sqreunnltb 27705. (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)))
Theorem	2sqreulem2 27696	Lemma 2 for 2sqreu 27700 etc. (Contributed by AV, 25-Jun-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ 𝐶 ∈ ℕ0) → (((𝐴↑2) + (𝐵↑2)) = ((𝐴↑2) + (𝐶↑2)) → 𝐵 = 𝐶))
Theorem	2sqreulem3 27697	Lemma 3 for 2sqreu 27700 etc. (Contributed by AV, 25-Jun-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ (𝐵 ∈ ℕ0 ∧ 𝐶 ∈ ℕ0)) → (((𝜑 ∧ ((𝐴↑2) + (𝐵↑2)) = 𝑃) ∧ (𝜓 ∧ ((𝐴↑2) + (𝐶↑2)) = 𝑃)) → 𝐵 = 𝐶))
Theorem	2sqreulem4 27698*	Lemma 4 for 2sqreu 27700 et. (Contributed by AV, 25-Jun-2023.)
⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))    ⇒   ⊢ ∀𝑎 ∈ ℕ0 ∃*𝑏 ∈ ℕ0 𝜑
Theorem	2sqreunnlem2 27699*	Lemma 2 for 2sqreunn 27701. (Contributed by AV, 25-Jun-2023.)
⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))    ⇒   ⊢ ∀𝑎 ∈ ℕ ∃*𝑏 ∈ ℕ 𝜑
Theorem	2sqreu 27700*	There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. See 2sqnn0 27682 for the existence of such a decomposition. (Contributed by AV, 4-Jun-2023.) (Revised by AV, 25-Jun-2023.)
⊢ (𝜑 ↔ (𝑎 ≤ 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃))    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → (∃!𝑎 ∈ ℕ0 ∃𝑏 ∈ ℕ0 𝜑 ∧ ∃!𝑏 ∈ ℕ0 ∃𝑎 ∈ ℕ0 𝜑))