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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | gausslemma2d 27301* | 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 27302* | Lemma for lgseisen 27306. 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 27303* | Lemma for lgseisen 27306. 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 27304* | Lemma for lgseisen 27306. (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 27305* | Lemma for lgseisen 27306. (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 27306* | 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 27307* | Lemma for lgsquad 27310. 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 27308* | Lemma for lgsquad 27310. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27307 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 27309* | Lemma for lgsquad 27310. (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 27310 | 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 27311 | Lemma for lgsquad2 27313. (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 27312* | Lemma for lgsquad2 27313. (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 27313 | Extend lgsquad 27310 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 27314 | Extend lgsquad2 27313 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 27315 | 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 27316* | Lemma 1 for 2lgslem1a 27318. (Contributed by AV, 16-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃)) | ||
| Theorem | 2lgslem1a2 27317 | Lemma 2 for 2lgslem1a 27318. (Contributed by AV, 18-Jun-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2))) | ||
| Theorem | 2lgslem1a 27318* | Lemma 1 for 2lgslem1 27321. (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 27319* | Lemma 2 for 2lgslem1 27321. (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 27320 | Lemma 3 for 2lgslem1 27321. (Contributed by AV, 19-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2)) | ||
| Theorem | 2lgslem1 27321* | Lemma 1 for 2lgs 27334. (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 27322 | Lemma 2 for 2lgs 27334. (Contributed by AV, 20-Jun-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ) | ||
| Theorem | 2lgslem3a 27323 | Lemma for 2lgslem3a1 27327. (Contributed by AV, 14-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾)) | ||
| Theorem | 2lgslem3b 27324 | Lemma for 2lgslem3b1 27328. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3c 27325 | Lemma for 2lgslem3c1 27329. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3d 27326 | Lemma for 2lgslem3d1 27330. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2)) | ||
| Theorem | 2lgslem3a1 27327 | Lemma 1 for 2lgslem3 27331. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3b1 27328 | Lemma 2 for 2lgslem3 27331. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3c1 27329 | Lemma 3 for 2lgslem3 27331. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3d1 27330 | Lemma 4 for 2lgslem3 27331. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3 27331 | Lemma 3 for 2lgs 27334. (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 27332 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
| ⊢ (2 /L 2) = 0 | ||
| Theorem | 2lgslem4 27333 | Lemma 4 for 2lgs 27334: special case of 2lgs 27334 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.) |
| ⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7}) | ||
| Theorem | 2lgs 27334 | 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 27241) 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 27335 | Lemma 1 for 2lgsoddprm 27343. (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 27336 | Lemma 2 for 2lgsoddprm 27343. (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 27337 | Lemma 1 for 2lgsoddprmlem3 27341. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((1↑2) − 1) / 8) = 0 | ||
| Theorem | 2lgsoddprmlem3b 27338 | Lemma 2 for 2lgsoddprmlem3 27341. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((3↑2) − 1) / 8) = 1 | ||
| Theorem | 2lgsoddprmlem3c 27339 | Lemma 3 for 2lgsoddprmlem3 27341. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((5↑2) − 1) / 8) = 3 | ||
| Theorem | 2lgsoddprmlem3d 27340 | Lemma 4 for 2lgsoddprmlem3 27341. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((7↑2) − 1) / 8) = (2 · 3) | ||
| Theorem | 2lgsoddprmlem3 27341 | Lemma 3 for 2lgsoddprm 27343. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7})) | ||
| Theorem | 2lgsoddprmlem4 27342 | Lemma 4 for 2lgsoddprm 27343. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7})) | ||
| Theorem | 2lgsoddprm 27343 | 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 27344* | Lemma for 2sq 27357. (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 27345* | Lemma for 2sq 27357. (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 27346 | 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 27347 | Lemma for 2sqlem5 27349. (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 27348 | Lemma for 2sqlem5 27349. (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 27349 | Lemma for 2sq 27357. 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 27350* | Lemma for 2sq 27357. 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 27351* | Lemma for 2sq 27357. (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 27352* | Lemma for 2sqlem8 27353. (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 27353* | Lemma for 2sq 27357. (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 27354* | Lemma for 2sq 27357. (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 27355* | Lemma for 2sq 27357. 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 27356* | Lemma for 2sq 27357. (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 27357* | 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 27358 | Lemma for 2sqb 27359. (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 27359* | The converse to 2sq 27357. (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 27360 | 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 27361 | 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 27362 | 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 27363 | 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 27364* | There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27359 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 27365* | 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 27366* | 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 27367* |
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 27370, 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 2562 and 2eu4 2648. For more details see comment for addsqnreup 27370. (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqn2reu 27368* |
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 27367, 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 27369* |
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 27367, 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 27367). For more details see comment for addsqnreup 27370. (Contributed by AV, 20-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqnreup 27370* |
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 27367, 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 2562 and 2eu4 2648. 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 27367). 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 27369). 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 4476 resp. 2eu4 2648). These two representations are equivalent (see opreu2reurex 6246). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27372. 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 27388 and 2sqreuopb 27395). 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 48774). (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶) | ||
| Theorem | addsq2nreurex 27371* | 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 27372* |
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 27367, 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 2562 and 2eu4 2648. (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 27373* | Lemma 1 for 2sqreu 27383. (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 27374* | Lemma for 2sqreult 27385. (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 27375* | Lemma for 2sqreultb 27386. (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 27376* | Lemma 1 for 2sqreunn 27384. (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 27377* | Lemma for 2sqreunnlt 27387. (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 27378* | Lemma for 2sqreunnltb 27388. (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 27379 | Lemma 2 for 2sqreu 27383 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 27380 | Lemma 3 for 2sqreu 27383 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 27381* | Lemma 4 for 2sqreu 27383 et. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ0 ∃*𝑏 ∈ ℕ0 𝜑 | ||
| Theorem | 2sqreunnlem2 27382* | Lemma 2 for 2sqreunn 27384. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ ∃*𝑏 ∈ ℕ 𝜑 | ||
| Theorem | 2sqreu 27383* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. See 2sqnn0 27365 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 27384* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two positive integers. See 2sqnn 27366 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 27385* | 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 27386* | 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 27387* | 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 27388* | 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 27389* | 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 27383. (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 27390* | 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 27384. (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 27391* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers. Ordered pair variant of 2sqreult 27385. (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 27392* | 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 27386. (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 27393* | 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 27387. (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 27394* | 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 27388. (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 | 2sqreuopb 27395* | 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. Alternate ordered pair variant of 2sqreunnltb 27388. (Contributed by AV, 3-Jul-2023.) |
| ⊢ (𝑃 ∈ ℙ → ((𝑃 mod 4) = 1 ↔ ∃!𝑝 ∈ (ℕ × ℕ)∃𝑎∃𝑏(𝑝 = 〈𝑎, 𝑏〉 ∧ (𝑎 < 𝑏 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)))) | ||
| Theorem | chebbnd1lem1 27396 | Lemma for chebbnd1 27399: show a lower bound on π(𝑥) at even integers using similar techniques to those used to prove bpos 27220. (Note that the expression 𝐾 is actually equal to 2 · 𝑁, but proving that is not necessary for the proof, and it's too much work.) The key to the proof is bposlem1 27211, which shows that each term in the expansion ((2 · 𝑁)C𝑁) = ∏𝑝 ∈ ℙ (𝑝↑(𝑝 pCnt ((2 · 𝑁)C𝑁))) is at most 2 · 𝑁, so that the sum really only has nonzero elements up to 2 · 𝑁, and since each term is at most 2 · 𝑁, after taking logs we get the inequality π(2 · 𝑁) · log(2 · 𝑁) ≤ log((2 · 𝑁)C𝑁), and bclbnd 27207 finishes the proof. (Contributed by Mario Carneiro, 22-Sep-2014.) (Revised by Mario Carneiro, 15-Apr-2016.) |
| ⊢ 𝐾 = if((2 · 𝑁) ≤ ((2 · 𝑁)C𝑁), (2 · 𝑁), ((2 · 𝑁)C𝑁)) ⇒ ⊢ (𝑁 ∈ (ℤ≥‘4) → (log‘((4↑𝑁) / 𝑁)) < ((π‘(2 · 𝑁)) · (log‘(2 · 𝑁)))) | ||
| Theorem | chebbnd1lem2 27397 | Lemma for chebbnd1 27399: Show that log(𝑁) / 𝑁 does not change too much between 𝑁 and 𝑀 = ⌊(𝑁 / 2). (Contributed by Mario Carneiro, 22-Sep-2014.) |
| ⊢ 𝑀 = (⌊‘(𝑁 / 2)) ⇒ ⊢ ((𝑁 ∈ ℝ ∧ 8 ≤ 𝑁) → ((log‘(2 · 𝑀)) / (2 · 𝑀)) < (2 · ((log‘𝑁) / 𝑁))) | ||
| Theorem | chebbnd1lem3 27398 | Lemma for chebbnd1 27399: get a lower bound on π(𝑁) / (𝑁 / log(𝑁)) that is independent of 𝑁. (Contributed by Mario Carneiro, 21-Sep-2014.) |
| ⊢ 𝑀 = (⌊‘(𝑁 / 2)) ⇒ ⊢ ((𝑁 ∈ ℝ ∧ 8 ≤ 𝑁) → (((log‘2) − (1 / (2 · e))) / 2) < ((π‘𝑁) · ((log‘𝑁) / 𝑁))) | ||
| Theorem | chebbnd1 27399 | The Chebyshev bound: The function π(𝑥) is eventually lower bounded by a positive constant times 𝑥 / log(𝑥). Alternatively stated, the function (𝑥 / log(𝑥)) / π(𝑥) is eventually bounded. (Contributed by Mario Carneiro, 22-Sep-2014.) |
| ⊢ (𝑥 ∈ (2[,)+∞) ↦ ((𝑥 / (log‘𝑥)) / (π‘𝑥))) ∈ 𝑂(1) | ||
| Theorem | chtppilimlem1 27400 | Lemma for chtppilim 27402. (Contributed by Mario Carneiro, 22-Sep-2014.) |
| ⊢ (𝜑 → 𝐴 ∈ ℝ+) & ⊢ (𝜑 → 𝐴 < 1) & ⊢ (𝜑 → 𝑁 ∈ (2[,)+∞)) & ⊢ (𝜑 → ((𝑁↑𝑐𝐴) / (π‘𝑁)) < (1 − 𝐴)) ⇒ ⊢ (𝜑 → ((𝐴↑2) · ((π‘𝑁) · (log‘𝑁))) < (θ‘𝑁)) | ||
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