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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | gausslemma2dlem1a 27301* | Lemma for gausslemma2dlem1 27302. (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 27302* | Lemma 1 for gausslemma2d 27310. (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 27303* | Lemma 2 for gausslemma2d 27310. (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 27304* | Lemma 3 for gausslemma2d 27310. (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 27305* | Lemma 4 for gausslemma2d 27310. (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 27306* | Lemma for gausslemma2dlem5 27307. (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 27307* | Lemma 5 for gausslemma2d 27310. (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 27308* | Lemma 6 for gausslemma2d 27310. (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 27309* | Lemma 7 for gausslemma2d 27310. (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 27310* | 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)) | ||
| Theorem | lgseisenlem1 27311* | Lemma for lgseisen 27315. 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 27312* | Lemma for lgseisen 27315. 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 27313* | Lemma for lgseisen 27315. (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 27314* | Lemma for lgseisen 27315. (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 27315* | 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 27316* | Lemma for lgsquad 27319. 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 27317* | Lemma for lgsquad 27319. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27316 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 27318* | Lemma for lgsquad 27319. (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 27319 | 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 27320 | Lemma for lgsquad2 27322. (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 27321* | Lemma for lgsquad2 27322. (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 27322 | Extend lgsquad 27319 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 27323 | Extend lgsquad2 27322 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 27324 | 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 27325* | Lemma 1 for 2lgslem1a 27327. (Contributed by AV, 16-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃)) | ||
| Theorem | 2lgslem1a2 27326 | Lemma 2 for 2lgslem1a 27327. (Contributed by AV, 18-Jun-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2))) | ||
| Theorem | 2lgslem1a 27327* | Lemma 1 for 2lgslem1 27330. (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 27328* | Lemma 2 for 2lgslem1 27330. (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 27329 | Lemma 3 for 2lgslem1 27330. (Contributed by AV, 19-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2)) | ||
| Theorem | 2lgslem1 27330* | Lemma 1 for 2lgs 27343. (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 27331 | Lemma 2 for 2lgs 27343. (Contributed by AV, 20-Jun-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ) | ||
| Theorem | 2lgslem3a 27332 | Lemma for 2lgslem3a1 27336. (Contributed by AV, 14-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾)) | ||
| Theorem | 2lgslem3b 27333 | Lemma for 2lgslem3b1 27337. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3c 27334 | Lemma for 2lgslem3c1 27338. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3d 27335 | Lemma for 2lgslem3d1 27339. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2)) | ||
| Theorem | 2lgslem3a1 27336 | Lemma 1 for 2lgslem3 27340. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3b1 27337 | Lemma 2 for 2lgslem3 27340. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3c1 27338 | Lemma 3 for 2lgslem3 27340. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3d1 27339 | Lemma 4 for 2lgslem3 27340. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3 27340 | Lemma 3 for 2lgs 27343. (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 27341 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
| ⊢ (2 /L 2) = 0 | ||
| Theorem | 2lgslem4 27342 | Lemma 4 for 2lgs 27343: special case of 2lgs 27343 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.) |
| ⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7}) | ||
| Theorem | 2lgs 27343 | 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 27250) 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 27344 | Lemma 1 for 2lgsoddprm 27352. (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 27345 | Lemma 2 for 2lgsoddprm 27352. (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 27346 | Lemma 1 for 2lgsoddprmlem3 27350. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((1↑2) − 1) / 8) = 0 | ||
| Theorem | 2lgsoddprmlem3b 27347 | Lemma 2 for 2lgsoddprmlem3 27350. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((3↑2) − 1) / 8) = 1 | ||
| Theorem | 2lgsoddprmlem3c 27348 | Lemma 3 for 2lgsoddprmlem3 27350. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((5↑2) − 1) / 8) = 3 | ||
| Theorem | 2lgsoddprmlem3d 27349 | Lemma 4 for 2lgsoddprmlem3 27350. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((7↑2) − 1) / 8) = (2 · 3) | ||
| Theorem | 2lgsoddprmlem3 27350 | Lemma 3 for 2lgsoddprm 27352. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7})) | ||
| Theorem | 2lgsoddprmlem4 27351 | Lemma 4 for 2lgsoddprm 27352. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7})) | ||
| Theorem | 2lgsoddprm 27352 | 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))) | ||
| Theorem | 2sqlem1 27353* | Lemma for 2sq 27366. (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 27354* | Lemma for 2sq 27366. (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 27355 | 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 27356 | Lemma for 2sqlem5 27358. (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 27357 | Lemma for 2sqlem5 27358. (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 27358 | Lemma for 2sq 27366. 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 27359* | Lemma for 2sq 27366. 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 27360* | Lemma for 2sq 27366. (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 27361* | Lemma for 2sqlem8 27362. (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 27362* | Lemma for 2sq 27366. (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 27363* | Lemma for 2sq 27366. (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 27364* | Lemma for 2sq 27366. 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 27365* | Lemma for 2sq 27366. (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 27366* | 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 27367 | Lemma for 2sqb 27368. (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 27368* | The converse to 2sq 27366. (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 27369 | 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 27370 | 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 27371 | 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 27372 | 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 27373* | There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27368 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 27374* | 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 27375* | 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 27376* |
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 27379, 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 2564 and 2eu4 2650. For more details see comment for addsqnreup 27379. (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqn2reu 27377* |
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 27376, 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 27378* |
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 27376, 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 27376). For more details see comment for addsqnreup 27379. (Contributed by AV, 20-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqnreup 27379* |
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 27376, 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 2564 and 2eu4 2650. 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 27376). 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 27378). 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 4473 resp. 2eu4 2650). These two representations are equivalent (see opreu2reurex 6241). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27381. 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 27397 and 2sqreuopb 27404). 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 48819). (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶) | ||
| Theorem | addsq2nreurex 27380* | 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 27381* |
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 27376, 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 2564 and 2eu4 2650. (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 27382* | Lemma 1 for 2sqreu 27392. (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 27383* | Lemma for 2sqreult 27394. (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 27384* | Lemma for 2sqreultb 27395. (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 27385* | Lemma 1 for 2sqreunn 27393. (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 27386* | Lemma for 2sqreunnlt 27396. (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 27387* | Lemma for 2sqreunnltb 27397. (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 27388 | Lemma 2 for 2sqreu 27392 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 27389 | Lemma 3 for 2sqreu 27392 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 27390* | Lemma 4 for 2sqreu 27392 et. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ0 ∃*𝑏 ∈ ℕ0 𝜑 | ||
| Theorem | 2sqreunnlem2 27391* | Lemma 2 for 2sqreunn 27393. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ ∃*𝑏 ∈ ℕ 𝜑 | ||
| Theorem | 2sqreu 27392* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. See 2sqnn0 27374 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)) | ||
| Theorem | 2sqreunn 27393* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two positive integers. See 2sqnn 27375 for the existence of such a decomposition. (Contributed by AV, 11-Jun-2023.) (Revised by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝑎 ≤ 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → (∃!𝑎 ∈ ℕ ∃𝑏 ∈ ℕ 𝜑 ∧ ∃!𝑏 ∈ ℕ ∃𝑎 ∈ ℕ 𝜑)) | ||
| Theorem | 2sqreult 27394* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers. (Contributed by AV, 8-Jun-2023.) (Proposed by GL, 8-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)) | ||
| Theorem | 2sqreultb 27395* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers iff 𝑃≡1 (mod 4). (Contributed by AV, 10-Jun-2023.) The prime needs not be odd, as observed by WL. (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))) | ||
| Theorem | 2sqreunnlt 27396* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two different positive integers. (Contributed by AV, 4-Jun-2023.) Specialization to different integers, proposed by GL. (Revised by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → (∃!𝑎 ∈ ℕ ∃𝑏 ∈ ℕ 𝜑 ∧ ∃!𝑏 ∈ ℕ ∃𝑎 ∈ ℕ 𝜑)) | ||
| Theorem | 2sqreunnltb 27397* | There exists a unique decomposition of a prime as a sum of squares of two different positive integers iff the prime is of the form 4𝑘 + 1. (Contributed by AV, 11-Jun-2023.) The prime needs not be odd, as observed by WL. (Revised by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ (𝑃 ∈ ℙ → ((𝑃 mod 4) = 1 ↔ (∃!𝑎 ∈ ℕ ∃𝑏 ∈ ℕ 𝜑 ∧ ∃!𝑏 ∈ ℕ ∃𝑎 ∈ ℕ 𝜑))) | ||
| Theorem | 2sqreuop 27398* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. Ordered pair variant of 2sqreu 27392. (Contributed by AV, 2-Jul-2023.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑝 ∈ (ℕ0 × ℕ0)((1st ‘𝑝) ≤ (2nd ‘𝑝) ∧ (((1st ‘𝑝)↑2) + ((2nd ‘𝑝)↑2)) = 𝑃)) | ||
| Theorem | 2sqreuopnn 27399* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two positive integers. Ordered pair variant of 2sqreunn 27393. (Contributed by AV, 2-Jul-2023.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑝 ∈ (ℕ × ℕ)((1st ‘𝑝) ≤ (2nd ‘𝑝) ∧ (((1st ‘𝑝)↑2) + ((2nd ‘𝑝)↑2)) = 𝑃)) | ||
| Theorem | 2sqreuoplt 27400* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers. Ordered pair variant of 2sqreult 27394. (Contributed by AV, 2-Jul-2023.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑝 ∈ (ℕ0 × ℕ0)((1st ‘𝑝) < (2nd ‘𝑝) ∧ (((1st ‘𝑝)↑2) + ((2nd ‘𝑝)↑2)) = 𝑃)) | ||
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