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Type | Label | Description |
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Statement | ||
Theorem | lgsquadlem1 27401* | Lemma for lgsquad 27404. 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 27402* | Lemma for lgsquad 27404. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27401 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 27403* | Lemma for lgsquad 27404. (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 27404 | 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 27405 | Lemma for lgsquad2 27407. (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 27406* | Lemma for lgsquad2 27407. (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 27407 | Extend lgsquad 27404 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 27408 | Extend lgsquad2 27407 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 27409 | 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 27410* | Lemma 1 for 2lgslem1a 27412. (Contributed by AV, 16-Jun-2021.) |
⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃)) | ||
Theorem | 2lgslem1a2 27411 | Lemma 2 for 2lgslem1a 27412. (Contributed by AV, 18-Jun-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2))) | ||
Theorem | 2lgslem1a 27412* | Lemma 1 for 2lgslem1 27415. (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 27413* | Lemma 2 for 2lgslem1 27415. (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 27414 | Lemma 3 for 2lgslem1 27415. (Contributed by AV, 19-Jun-2021.) |
⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2)) | ||
Theorem | 2lgslem1 27415* | Lemma 1 for 2lgs 27428. (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 27416 | Lemma 2 for 2lgs 27428. (Contributed by AV, 20-Jun-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ) | ||
Theorem | 2lgslem3a 27417 | Lemma for 2lgslem3a1 27421. (Contributed by AV, 14-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾)) | ||
Theorem | 2lgslem3b 27418 | Lemma for 2lgslem3b1 27422. (Contributed by AV, 16-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
Theorem | 2lgslem3c 27419 | Lemma for 2lgslem3c1 27423. (Contributed by AV, 16-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
Theorem | 2lgslem3d 27420 | Lemma for 2lgslem3d1 27424. (Contributed by AV, 16-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2)) | ||
Theorem | 2lgslem3a1 27421 | Lemma 1 for 2lgslem3 27425. (Contributed by AV, 15-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0) | ||
Theorem | 2lgslem3b1 27422 | Lemma 2 for 2lgslem3 27425. (Contributed by AV, 16-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1) | ||
Theorem | 2lgslem3c1 27423 | Lemma 3 for 2lgslem3 27425. (Contributed by AV, 16-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1) | ||
Theorem | 2lgslem3d1 27424 | Lemma 4 for 2lgslem3 27425. (Contributed by AV, 15-Jul-2021.) |
⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0) | ||
Theorem | 2lgslem3 27425 | Lemma 3 for 2lgs 27428. (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 27426 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
⊢ (2 /L 2) = 0 | ||
Theorem | 2lgslem4 27427 | Lemma 4 for 2lgs 27428: special case of 2lgs 27428 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.) |
⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7}) | ||
Theorem | 2lgs 27428 | 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 27335) 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 27429 | Lemma 1 for 2lgsoddprm 27437. (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 27430 | Lemma 2 for 2lgsoddprm 27437. (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 27431 | Lemma 1 for 2lgsoddprmlem3 27435. (Contributed by AV, 20-Jul-2021.) |
⊢ (((1↑2) − 1) / 8) = 0 | ||
Theorem | 2lgsoddprmlem3b 27432 | Lemma 2 for 2lgsoddprmlem3 27435. (Contributed by AV, 20-Jul-2021.) |
⊢ (((3↑2) − 1) / 8) = 1 | ||
Theorem | 2lgsoddprmlem3c 27433 | Lemma 3 for 2lgsoddprmlem3 27435. (Contributed by AV, 20-Jul-2021.) |
⊢ (((5↑2) − 1) / 8) = 3 | ||
Theorem | 2lgsoddprmlem3d 27434 | Lemma 4 for 2lgsoddprmlem3 27435. (Contributed by AV, 20-Jul-2021.) |
⊢ (((7↑2) − 1) / 8) = (2 · 3) | ||
Theorem | 2lgsoddprmlem3 27435 | Lemma 3 for 2lgsoddprm 27437. (Contributed by AV, 20-Jul-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7})) | ||
Theorem | 2lgsoddprmlem4 27436 | Lemma 4 for 2lgsoddprm 27437. (Contributed by AV, 20-Jul-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7})) | ||
Theorem | 2lgsoddprm 27437 | 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 27438* | Lemma for 2sq 27451. (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 27439* | Lemma for 2sq 27451. (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 27440 | 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 27441 | Lemma for 2sqlem5 27443. (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 27442 | Lemma for 2sqlem5 27443. (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 27443 | Lemma for 2sq 27451. 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 27444* | Lemma for 2sq 27451. 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 27445* | Lemma for 2sq 27451. (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 27446* | Lemma for 2sqlem8 27447. (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 27447* | Lemma for 2sq 27451. (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 27448* | Lemma for 2sq 27451. (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 27449* | Lemma for 2sq 27451. 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 27450* | Lemma for 2sq 27451. (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 27451* | 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 27452 | Lemma for 2sqb 27453. (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 27453* | The converse to 2sq 27451. (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 27454 | 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 27455 | 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 27456 | 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 27457 | 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 27458* | There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27453 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 27459* | 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 27460* | 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 27461* |
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 27464, 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 2557 and 2eu4 2643. For more details see comment for addsqnreup 27464. (Contributed by AV, 21-Jun-2023.) |
⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
Theorem | addsqn2reu 27462* |
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 27461, 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 27463* |
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 27461, 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 27461). For more details see comment for addsqnreup 27464. (Contributed by AV, 20-Jun-2023.) |
⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
Theorem | addsqnreup 27464* |
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 27461, 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 2557 and 2eu4 2643. 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 27461). 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 27463). 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 4530 resp. 2eu4 2643). These two representations are equivalent (see opreu2reurex 6304). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27466. 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 27482 and 2sqreuopb 27489). 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 48113). (Contributed by AV, 21-Jun-2023.) |
⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶) | ||
Theorem | addsq2nreurex 27465* | 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 27466* |
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 27461, 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 2557 and 2eu4 2643. (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 27467* | Lemma 1 for 2sqreu 27477. (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 27468* | Lemma for 2sqreult 27479. (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 27469* | Lemma for 2sqreultb 27480. (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 27470* | Lemma 1 for 2sqreunn 27478. (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 27471* | Lemma for 2sqreunnlt 27481. (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 27472* | Lemma for 2sqreunnltb 27482. (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 27473 | Lemma 2 for 2sqreu 27477 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 27474 | Lemma 3 for 2sqreu 27477 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 27475* | Lemma 4 for 2sqreu 27477 et. (Contributed by AV, 25-Jun-2023.) |
⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ0 ∃*𝑏 ∈ ℕ0 𝜑 | ||
Theorem | 2sqreunnlem2 27476* | Lemma 2 for 2sqreunn 27478. (Contributed by AV, 25-Jun-2023.) |
⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ ∃*𝑏 ∈ ℕ 𝜑 | ||
Theorem | 2sqreu 27477* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. See 2sqnn0 27459 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 27478* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two positive integers. See 2sqnn 27460 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 27479* | 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 27480* | 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 27481* | 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 27482* | 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 27483* | 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 27477. (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 27484* | 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 27478. (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 27485* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers. Ordered pair variant of 2sqreult 27479. (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 27486* | 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 27480. (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 27487* | 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 27481. (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 27488* | 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 27482. (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 27489* | 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 27482. (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 27490 | Lemma for chebbnd1 27493: show a lower bound on π(𝑥) at even integers using similar techniques to those used to prove bpos 27314. (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 27305, 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 27301 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 27491 | Lemma for chebbnd1 27493: 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 27492 | Lemma for chebbnd1 27493: 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 27493 | 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 27494 | Lemma for chtppilim 27496. (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)) | ||
Theorem | chtppilimlem2 27495* | Lemma for chtppilim 27496. (Contributed by Mario Carneiro, 22-Sep-2014.) |
⊢ (𝜑 → 𝐴 ∈ ℝ+) & ⊢ (𝜑 → 𝐴 < 1) ⇒ ⊢ (𝜑 → ∃𝑧 ∈ ℝ ∀𝑥 ∈ (2[,)+∞)(𝑧 ≤ 𝑥 → ((𝐴↑2) · ((π‘𝑥) · (log‘𝑥))) < (θ‘𝑥))) | ||
Theorem | chtppilim 27496 | The θ function is asymptotic to π(𝑥)log(𝑥), so it is sufficient to prove θ(𝑥) / 𝑥 ⇝𝑟 1 to establish the PNT. (Contributed by Mario Carneiro, 22-Sep-2014.) |
⊢ (𝑥 ∈ (2[,)+∞) ↦ ((θ‘𝑥) / ((π‘𝑥) · (log‘𝑥)))) ⇝𝑟 1 | ||
Theorem | chto1ub 27497 | The θ function is upper bounded by a linear term. Corollary of chtub 27233. (Contributed by Mario Carneiro, 22-Sep-2014.) |
⊢ (𝑥 ∈ ℝ+ ↦ ((θ‘𝑥) / 𝑥)) ∈ 𝑂(1) | ||
Theorem | chebbnd2 27498 | The Chebyshev bound, part 2: The function π(𝑥) is eventually upper bounded by a positive constant times 𝑥 / log(𝑥). Alternatively stated, the function π(𝑥) / (𝑥 / log(𝑥)) is eventually bounded. (Contributed by Mario Carneiro, 22-Sep-2014.) |
⊢ (𝑥 ∈ (2[,)+∞) ↦ ((π‘𝑥) / (𝑥 / (log‘𝑥)))) ∈ 𝑂(1) | ||
Theorem | chto1lb 27499 | The θ function is lower bounded by a linear term. Corollary of chebbnd1 27493. (Contributed by Mario Carneiro, 8-Apr-2016.) |
⊢ (𝑥 ∈ (2[,)+∞) ↦ (𝑥 / (θ‘𝑥))) ∈ 𝑂(1) | ||
Theorem | chpchtlim 27500 | The ψ and θ functions are asymptotic to each other, so is sufficient to prove either θ(𝑥) / 𝑥 ⇝𝑟 1 or ψ(𝑥) / 𝑥 ⇝𝑟 1 to establish the PNT. (Contributed by Mario Carneiro, 8-Apr-2016.) |
⊢ (𝑥 ∈ (2[,)+∞) ↦ ((ψ‘𝑥) / (θ‘𝑥))) ⇝𝑟 1 |
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