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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | gausslemma2dlem3 27701* | Lemma 3 for gausslemma2d 27707. (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 27702* | Lemma 4 for gausslemma2d 27707. (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 27703* | Lemma for gausslemma2dlem5 27704. (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 27704* | Lemma 5 for gausslemma2d 27707. (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 27705* | Lemma 6 for gausslemma2d 27707. (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 27706* | Lemma 7 for gausslemma2d 27707. (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 27707* | 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 27708* | Lemma for lgseisen 27712. 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 27709* | Lemma for lgseisen 27712. 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 27710* | Lemma for lgseisen 27712. (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 27711* | Lemma for lgseisen 27712. (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 27712* | 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 27713* | Lemma for lgsquad 27716. 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 27714* | Lemma for lgsquad 27716. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27713 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 27715* | Lemma for lgsquad 27716. (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 27716 | 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 27717 | Lemma for lgsquad2 27719. (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 27718* | Lemma for lgsquad2 27719. (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 27719 | Extend lgsquad 27716 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 27720 | Extend lgsquad2 27719 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 27721 | 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 27722* | Lemma 1 for 2lgslem1a 27724. (Contributed by AV, 16-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃)) | ||
| Theorem | 2lgslem1a2 27723 | Lemma 2 for 2lgslem1a 27724. (Contributed by AV, 18-Jun-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2))) | ||
| Theorem | 2lgslem1a 27724* | Lemma 1 for 2lgslem1 27727. (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 27725* | Lemma 2 for 2lgslem1 27727. (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 27726 | Lemma 3 for 2lgslem1 27727. (Contributed by AV, 19-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2)) | ||
| Theorem | 2lgslem1 27727* | Lemma 1 for 2lgs 27740. (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 27728 | Lemma 2 for 2lgs 27740. (Contributed by AV, 20-Jun-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ) | ||
| Theorem | 2lgslem3a 27729 | Lemma for 2lgslem3a1 27733. (Contributed by AV, 14-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾)) | ||
| Theorem | 2lgslem3b 27730 | Lemma for 2lgslem3b1 27734. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3c 27731 | Lemma for 2lgslem3c1 27735. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3d 27732 | Lemma for 2lgslem3d1 27736. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2)) | ||
| Theorem | 2lgslem3a1 27733 | Lemma 1 for 2lgslem3 27737. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3b1 27734 | Lemma 2 for 2lgslem3 27737. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3c1 27735 | Lemma 3 for 2lgslem3 27737. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3d1 27736 | Lemma 4 for 2lgslem3 27737. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3 27737 | Lemma 3 for 2lgs 27740. (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 27738 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
| ⊢ (2 /L 2) = 0 | ||
| Theorem | 2lgslem4 27739 | Lemma 4 for 2lgs 27740: special case of 2lgs 27740 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.) |
| ⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7}) | ||
| Theorem | 2lgs 27740 | 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 27647) 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 27741 | Lemma 1 for 2lgsoddprm 27749. (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 27742 | Lemma 2 for 2lgsoddprm 27749. (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 27743 | Lemma 1 for 2lgsoddprmlem3 27747. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((1↑2) − 1) / 8) = 0 | ||
| Theorem | 2lgsoddprmlem3b 27744 | Lemma 2 for 2lgsoddprmlem3 27747. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((3↑2) − 1) / 8) = 1 | ||
| Theorem | 2lgsoddprmlem3c 27745 | Lemma 3 for 2lgsoddprmlem3 27747. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((5↑2) − 1) / 8) = 3 | ||
| Theorem | 2lgsoddprmlem3d 27746 | Lemma 4 for 2lgsoddprmlem3 27747. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((7↑2) − 1) / 8) = (2 · 3) | ||
| Theorem | 2lgsoddprmlem3 27747 | Lemma 3 for 2lgsoddprm 27749. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7})) | ||
| Theorem | 2lgsoddprmlem4 27748 | Lemma 4 for 2lgsoddprm 27749. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7})) | ||
| Theorem | 2lgsoddprm 27749 | 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 27750* | Lemma for 2sq 27763. (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 27751* | Lemma for 2sq 27763. (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 27752 | 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 27753 | Lemma for 2sqlem5 27755. (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 27754 | Lemma for 2sqlem5 27755. (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 27755 | Lemma for 2sq 27763. 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 27756* | Lemma for 2sq 27763. 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 27757* | Lemma for 2sq 27763. (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 27758* | Lemma for 2sqlem8 27759. (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 27759* | Lemma for 2sq 27763. (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 27760* | Lemma for 2sq 27763. (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 27761* | Lemma for 2sq 27763. 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 27762* | Lemma for 2sq 27763. (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 27763* | 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 27764 | Lemma for 2sqb 27765. (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 27765* | The converse to 2sq 27763. (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 27766 | 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 27767 | 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 27768 | 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 27769 | 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 27770* | There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27765 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 27771* | 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 27772* | 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 27773* |
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 27776, 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 2594 and 2eu4 2679. For more details see comment for addsqnreup 27776. (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqn2reu 27774* |
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 27773, 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 27775* |
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 27773, 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 27773). For more details see comment for addsqnreup 27776. (Contributed by AV, 20-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqnreup 27776* |
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 27773, 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 2594 and 2eu4 2679. 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 27773). 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 27775). 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 4479 resp. 2eu4 2679). These two representations are equivalent (see opreu2reurex 6297). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27778. 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 27794 and 2sqreuopb 27801). 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 49994). (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶) | ||
| Theorem | addsq2nreurex 27777* | 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 27778* |
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 27773, 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 2594 and 2eu4 2679. (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 27779* | Lemma 1 for 2sqreu 27789. (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 27780* | Lemma for 2sqreult 27791. (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 27781* | Lemma for 2sqreultb 27792. (Contributed by AV, 10-Jun-2023.) The prime need 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 27782* | Lemma 1 for 2sqreunn 27790. (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 27783* | Lemma for 2sqreunnlt 27793. (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 27784* | Lemma for 2sqreunnltb 27794. (Contributed by AV, 11-Jun-2023.) The prime need 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 27785 | Lemma 2 for 2sqreu 27789 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 27786 | Lemma 3 for 2sqreu 27789 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 27787* | Lemma 4 for 2sqreu 27789 et. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ0 ∃*𝑏 ∈ ℕ0 𝜑 | ||
| Theorem | 2sqreunnlem2 27788* | Lemma 2 for 2sqreunn 27790. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ ∃*𝑏 ∈ ℕ 𝜑 | ||
| Theorem | 2sqreu 27789* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. See 2sqnn0 27771 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 27790* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two positive integers. See 2sqnn 27772 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 27791* | 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 27792* | 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 need 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 27793* | 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 27794* | 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 need 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 27795* | 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 27789. (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 27796* | 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 27790. (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 27797* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers. Ordered pair variant of 2sqreult 27791. (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 | 2sqreuopltb 27798* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers iff 𝑃≡1 (mod 4). Ordered pair variant of 2sqreultb 27792. (Contributed by AV, 3-Jul-2023.) |
| ⊢ (𝑃 ∈ ℙ → ((𝑃 mod 4) = 1 ↔ ∃!𝑝 ∈ (ℕ0 × ℕ0)((1st ‘𝑝) < (2nd ‘𝑝) ∧ (((1st ‘𝑝)↑2) + ((2nd ‘𝑝)↑2)) = 𝑃))) | ||
| Theorem | 2sqreuopnnlt 27799* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two different positive integers. Ordered pair variant of 2sqreunnlt 27793. (Contributed by AV, 3-Jul-2023.) |
| ⊢ ((𝑃 ∈ ℙ ∧ (𝑃 mod 4) = 1) → ∃!𝑝 ∈ (ℕ × ℕ)((1st ‘𝑝) < (2nd ‘𝑝) ∧ (((1st ‘𝑝)↑2) + ((2nd ‘𝑝)↑2)) = 𝑃)) | ||
| Theorem | 2sqreuopnnltb 27800* | 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. Ordered pair variant of 2sqreunnltb 27794. (Contributed by AV, 3-Jul-2023.) |
| ⊢ (𝑃 ∈ ℙ → ((𝑃 mod 4) = 1 ↔ ∃!𝑝 ∈ (ℕ × ℕ)((1st ‘𝑝) < (2nd ‘𝑝) ∧ (((1st ‘𝑝)↑2) + ((2nd ‘𝑝)↑2)) = 𝑃))) | ||
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