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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | lgsquadlem2 27701* | Lemma for lgsquad 27703. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 27700 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 27702* | Lemma for lgsquad 27703. (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 27703 | 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 27704 | Lemma for lgsquad2 27706. (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 27705* | Lemma for lgsquad2 27706. (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 27706 | Extend lgsquad 27703 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 27707 | Extend lgsquad2 27706 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 27708 | 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 27709* | Lemma 1 for 2lgslem1a 27711. (Contributed by AV, 16-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃)) | ||
| Theorem | 2lgslem1a2 27710 | Lemma 2 for 2lgslem1a 27711. (Contributed by AV, 18-Jun-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2))) | ||
| Theorem | 2lgslem1a 27711* | Lemma 1 for 2lgslem1 27714. (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 27712* | Lemma 2 for 2lgslem1 27714. (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 27713 | Lemma 3 for 2lgslem1 27714. (Contributed by AV, 19-Jun-2021.) |
| ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2)) | ||
| Theorem | 2lgslem1 27714* | Lemma 1 for 2lgs 27727. (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 27715 | Lemma 2 for 2lgs 27727. (Contributed by AV, 20-Jun-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ) | ||
| Theorem | 2lgslem3a 27716 | Lemma for 2lgslem3a1 27720. (Contributed by AV, 14-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾)) | ||
| Theorem | 2lgslem3b 27717 | Lemma for 2lgslem3b1 27721. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3c 27718 | Lemma for 2lgslem3c1 27722. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1)) | ||
| Theorem | 2lgslem3d 27719 | Lemma for 2lgslem3d1 27723. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝐾 ∈ ℕ0 ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2)) | ||
| Theorem | 2lgslem3a1 27720 | Lemma 1 for 2lgslem3 27724. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3b1 27721 | Lemma 2 for 2lgslem3 27724. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3c1 27722 | Lemma 3 for 2lgslem3 27724. (Contributed by AV, 16-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1) | ||
| Theorem | 2lgslem3d1 27723 | Lemma 4 for 2lgslem3 27724. (Contributed by AV, 15-Jul-2021.) |
| ⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))) ⇒ ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0) | ||
| Theorem | 2lgslem3 27724 | Lemma 3 for 2lgs 27727. (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 27725 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
| ⊢ (2 /L 2) = 0 | ||
| Theorem | 2lgslem4 27726 | Lemma 4 for 2lgs 27727: special case of 2lgs 27727 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.) |
| ⊢ ((2 /L 2) = 1 ↔ (2 mod 8) ∈ {1, 7}) | ||
| Theorem | 2lgs 27727 | 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 27634) 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 27728 | Lemma 1 for 2lgsoddprm 27736. (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 27729 | Lemma 2 for 2lgsoddprm 27736. (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 27730 | Lemma 1 for 2lgsoddprmlem3 27734. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((1↑2) − 1) / 8) = 0 | ||
| Theorem | 2lgsoddprmlem3b 27731 | Lemma 2 for 2lgsoddprmlem3 27734. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((3↑2) − 1) / 8) = 1 | ||
| Theorem | 2lgsoddprmlem3c 27732 | Lemma 3 for 2lgsoddprmlem3 27734. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((5↑2) − 1) / 8) = 3 | ||
| Theorem | 2lgsoddprmlem3d 27733 | Lemma 4 for 2lgsoddprmlem3 27734. (Contributed by AV, 20-Jul-2021.) |
| ⊢ (((7↑2) − 1) / 8) = (2 · 3) | ||
| Theorem | 2lgsoddprmlem3 27734 | Lemma 3 for 2lgsoddprm 27736. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7})) | ||
| Theorem | 2lgsoddprmlem4 27735 | Lemma 4 for 2lgsoddprm 27736. (Contributed by AV, 20-Jul-2021.) |
| ⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7})) | ||
| Theorem | 2lgsoddprm 27736 | 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 27737* | Lemma for 2sq 27750. (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 27738* | Lemma for 2sq 27750. (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 27739 | 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 27740 | Lemma for 2sqlem5 27742. (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 27741 | Lemma for 2sqlem5 27742. (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 27742 | Lemma for 2sq 27750. 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 27743* | Lemma for 2sq 27750. 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 27744* | Lemma for 2sq 27750. (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 27745* | Lemma for 2sqlem8 27746. (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 27746* | Lemma for 2sq 27750. (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 27747* | Lemma for 2sq 27750. (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 27748* | Lemma for 2sq 27750. 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 27749* | Lemma for 2sq 27750. (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 27750* | 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 27751 | Lemma for 2sqb 27752. (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 27752* | The converse to 2sq 27750. (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 27753 | 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 27754 | 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 27755 | 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 27756 | 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 27757* | There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27752 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 27758* | 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 27759* | 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 27760* |
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 27763, 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 2595 and 2eu4 2680. For more details see comment for addsqnreup 27763. (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃!𝑎 ∈ ℂ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqn2reu 27761* |
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 27760, 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 27762* |
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 27760, 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 27760). For more details see comment for addsqnreup 27763. (Contributed by AV, 20-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ∃𝑎 ∈ ℂ ¬ ∃!𝑏 ∈ ℂ (𝑎 + (𝑏↑2)) = 𝐶) | ||
| Theorem | addsqnreup 27763* |
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 27760, 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 2595 and 2eu4 2680. 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 27760). 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 27762). 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 4480 resp. 2eu4 2680). These two representations are equivalent (see opreu2reurex 6296). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27765. 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 27781 and 2sqreuopb 27788). 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 49869). (Contributed by AV, 21-Jun-2023.) |
| ⊢ (𝐶 ∈ ℂ → ¬ ∃!𝑝 ∈ (ℂ × ℂ)((1st ‘𝑝) + ((2nd ‘𝑝)↑2)) = 𝐶) | ||
| Theorem | addsq2nreurex 27764* | 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 27765* |
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 27760, 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 2595 and 2eu4 2680. (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 27766* | Lemma 1 for 2sqreu 27776. (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 27767* | Lemma for 2sqreult 27778. (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 27768* | Lemma for 2sqreultb 27779. (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 27769* | Lemma 1 for 2sqreunn 27777. (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 27770* | Lemma for 2sqreunnlt 27780. (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 27771* | Lemma for 2sqreunnltb 27781. (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 27772 | Lemma 2 for 2sqreu 27776 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 27773 | Lemma 3 for 2sqreu 27776 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 27774* | Lemma 4 for 2sqreu 27776 et. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ0 ∃*𝑏 ∈ ℕ0 𝜑 | ||
| Theorem | 2sqreunnlem2 27775* | Lemma 2 for 2sqreunn 27777. (Contributed by AV, 25-Jun-2023.) |
| ⊢ (𝜑 ↔ (𝜓 ∧ ((𝑎↑2) + (𝑏↑2)) = 𝑃)) ⇒ ⊢ ∀𝑎 ∈ ℕ ∃*𝑏 ∈ ℕ 𝜑 | ||
| Theorem | 2sqreu 27776* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two nonnegative integers. See 2sqnn0 27758 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 27777* | There exists a unique decomposition of a prime of the form 4𝑘 + 1 as a sum of squares of two positive integers. See 2sqnn 27759 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 27778* | 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 27779* | 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 27780* | 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 27781* | 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 27782* | 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 27776. (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 27783* | 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 27777. (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 27784* | There exists a unique decomposition of a prime as a sum of squares of two different nonnegative integers. Ordered pair variant of 2sqreult 27778. (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 27785* | 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 27779. (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 27786* | 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 27780. (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 27787* | 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 27781. (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 27788* | 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 27781. (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 27789 | Lemma for chebbnd1 27792: show a lower bound on π(𝑥) at even integers using similar techniques to those used to prove bpos 27613. (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 27604, 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 27600 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 27790 | Lemma for chebbnd1 27792: 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 27791 | Lemma for chebbnd1 27792: 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 27792 | 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 27793 | Lemma for chtppilim 27795. (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 27794* | Lemma for chtppilim 27795. (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 27795 | 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 27796 | The θ function is upper bounded by a linear term. Corollary of chtub 27532. (Contributed by Mario Carneiro, 22-Sep-2014.) |
| ⊢ (𝑥 ∈ ℝ+ ↦ ((θ‘𝑥) / 𝑥)) ∈ 𝑂(1) | ||
| Theorem | chebbnd2 27797 | 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 27798 | The θ function is lower bounded by a linear term. Corollary of chebbnd1 27792. (Contributed by Mario Carneiro, 8-Apr-2016.) |
| ⊢ (𝑥 ∈ (2[,)+∞) ↦ (𝑥 / (θ‘𝑥))) ∈ 𝑂(1) | ||
| Theorem | chpchtlim 27799 | 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 | ||
| Theorem | chpo1ub 27800 | The ψ function is upper bounded by a linear term. (Contributed by Mario Carneiro, 16-Apr-2016.) |
| ⊢ (𝑥 ∈ ℝ+ ↦ ((ψ‘𝑥) / 𝑥)) ∈ 𝑂(1) | ||
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