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Type | Label | Description |
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Statement | ||
Theorem | lgsdchr 27301* | The Legendre symbol function π(π) = (π /L π), where π is an odd positive number, is a real Dirichlet character modulo π. (Contributed by Mario Carneiro, 28-Apr-2016.) |
β’ πΊ = (DChrβπ) & β’ π = (β€/nβ€βπ) & β’ π· = (BaseβπΊ) & β’ π΅ = (Baseβπ) & β’ πΏ = (β€RHomβπ) & β’ π = (π¦ β π΅ β¦ (β©ββπ β β€ (π¦ = (πΏβπ) β§ β = (π /L π)))) β β’ ((π β β β§ Β¬ 2 β₯ π) β (π β π· β§ π:π΅βΆβ)) | ||
Gauss' Lemma is valid for any integer not dividing the given prime number. In the following, only the special case for 2 (not dividing any odd prime) is proven, see gausslemma2d 27320. The general case is still to prove. | ||
Theorem | gausslemma2dlem0a 27302 | Auxiliary lemma 1 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) β β’ (π β π β β) | ||
Theorem | gausslemma2dlem0b 27303 | Auxiliary lemma 2 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) β β’ (π β π» β β) | ||
Theorem | gausslemma2dlem0c 27304 | Auxiliary lemma 3 for gausslemma2d 27320. (Contributed by AV, 13-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) β β’ (π β ((!βπ») gcd π) = 1) | ||
Theorem | gausslemma2dlem0d 27305 | Auxiliary lemma 4 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π = (ββ(π / 4)) β β’ (π β π β β0) | ||
Theorem | gausslemma2dlem0e 27306 | Auxiliary lemma 5 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π = (ββ(π / 4)) β β’ (π β (π Β· 2) < (π / 2)) | ||
Theorem | gausslemma2dlem0f 27307 | Auxiliary lemma 6 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π = (ββ(π / 4)) & β’ π» = ((π β 1) / 2) β β’ (π β (π + 1) β€ π») | ||
Theorem | gausslemma2dlem0g 27308 | Auxiliary lemma 7 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π = (ββ(π / 4)) & β’ π» = ((π β 1) / 2) β β’ (π β π β€ π») | ||
Theorem | gausslemma2dlem0h 27309 | Auxiliary lemma 8 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π = (ββ(π / 4)) & β’ π» = ((π β 1) / 2) & β’ π = (π» β π) β β’ (π β π β β0) | ||
Theorem | gausslemma2dlem0i 27310 | Auxiliary lemma 9 for gausslemma2d 27320. (Contributed by AV, 14-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π = (ββ(π / 4)) & β’ π» = ((π β 1) / 2) & β’ π = (π» β π) β β’ (π β (((2 /L π) mod π) = ((-1βπ) mod π) β (2 /L π) = (-1βπ))) | ||
Theorem | gausslemma2dlem1a 27311* | Lemma for gausslemma2dlem1 27312. (Contributed by AV, 1-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) β β’ (π β ran π = (1...π»)) | ||
Theorem | gausslemma2dlem1 27312* | Lemma 1 for gausslemma2d 27320. (Contributed by AV, 5-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) β β’ (π β (!βπ») = βπ β (1...π»)(π βπ)) | ||
Theorem | gausslemma2dlem2 27313* | Lemma 2 for gausslemma2d 27320. (Contributed by AV, 4-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) β β’ (π β βπ β (1...π)(π βπ) = (π Β· 2)) | ||
Theorem | gausslemma2dlem3 27314* | Lemma 3 for gausslemma2d 27320. (Contributed by AV, 4-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) β β’ (π β βπ β ((π + 1)...π»)(π βπ) = (π β (π Β· 2))) | ||
Theorem | gausslemma2dlem4 27315* | Lemma 4 for gausslemma2d 27320. (Contributed by AV, 16-Jun-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) β β’ (π β (!βπ») = (βπ β (1...π)(π βπ) Β· βπ β ((π + 1)...π»)(π βπ))) | ||
Theorem | gausslemma2dlem5a 27316* | Lemma for gausslemma2dlem5 27317. (Contributed by AV, 8-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) β β’ (π β (βπ β ((π + 1)...π»)(π βπ) mod π) = (βπ β ((π + 1)...π»)(-1 Β· (π Β· 2)) mod π)) | ||
Theorem | gausslemma2dlem5 27317* | Lemma 5 for gausslemma2d 27320. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) & β’ π = (π» β π) β β’ (π β (βπ β ((π + 1)...π»)(π βπ) mod π) = (((-1βπ) Β· βπ β ((π + 1)...π»)(π Β· 2)) mod π)) | ||
Theorem | gausslemma2dlem6 27318* | Lemma 6 for gausslemma2d 27320. (Contributed by AV, 16-Jun-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) & β’ π = (π» β π) β β’ (π β ((!βπ») mod π) = ((((-1βπ) Β· (2βπ»)) Β· (!βπ»)) mod π)) | ||
Theorem | gausslemma2dlem7 27319* | Lemma 7 for gausslemma2d 27320. (Contributed by AV, 13-Jul-2021.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) & β’ π = (π» β π) β β’ (π β (((-1βπ) Β· (2βπ»)) mod π) = 1) | ||
Theorem | gausslemma2d 27320* | 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.) |
β’ (π β π β (β β {2})) & β’ π» = ((π β 1) / 2) & β’ π = (π₯ β (1...π») β¦ if((π₯ Β· 2) < (π / 2), (π₯ Β· 2), (π β (π₯ Β· 2)))) & β’ π = (ββ(π / 4)) & β’ π = (π» β π) β β’ (π β (2 /L π) = (-1βπ)) | ||
Theorem | lgseisenlem1 27321* | Lemma for lgseisen 27325. If π (π’) = (π Β· π’) mod π and π(π’) = (-1βπ (π’)) Β· π (π’), then for any even 1 β€ π’ β€ π β 1, π(π’) is also an even integer 1 β€ π(π’) β€ π β 1. To simplify these statements, we divide all the even numbers by 2, so that it becomes the statement that π(π₯ / 2) = (-1βπ (π₯ / 2)) Β· π (π₯ / 2) / 2 is an integer between 1 and (π β 1) / 2. (Contributed by Mario Carneiro, 17-Jun-2015.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π Β· (2 Β· π₯)) mod π) & β’ π = (π₯ β (1...((π β 1) / 2)) β¦ ((((-1βπ ) Β· π ) mod π) / 2)) β β’ (π β π:(1...((π β 1) / 2))βΆ(1...((π β 1) / 2))) | ||
Theorem | lgseisenlem2 27322* | Lemma for lgseisen 27325. The function π is an injection (and hence a bijection by the pigeonhole principle). (Contributed by Mario Carneiro, 17-Jun-2015.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π Β· (2 Β· π₯)) mod π) & β’ π = (π₯ β (1...((π β 1) / 2)) β¦ ((((-1βπ ) Β· π ) mod π) / 2)) & β’ π = ((π Β· (2 Β· π¦)) mod π) β β’ (π β π:(1...((π β 1) / 2))β1-1-ontoβ(1...((π β 1) / 2))) | ||
Theorem | lgseisenlem3 27323* | Lemma for lgseisen 27325. (Contributed by Mario Carneiro, 17-Jun-2015.) (Proof shortened by AV, 28-Jul-2019.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π Β· (2 Β· π₯)) mod π) & β’ π = (π₯ β (1...((π β 1) / 2)) β¦ ((((-1βπ ) Β· π ) mod π) / 2)) & β’ π = ((π Β· (2 Β· π¦)) mod π) & β’ π = (β€/nβ€βπ) & β’ πΊ = (mulGrpβπ) & β’ πΏ = (β€RHomβπ) β β’ (π β (πΊ Ξ£g (π₯ β (1...((π β 1) / 2)) β¦ (πΏβ((-1βπ ) Β· π)))) = (1rβπ)) | ||
Theorem | lgseisenlem4 27324* | Lemma for lgseisen 27325. The function π is an injection (and hence a bijection by the pigeonhole principle). (Contributed by Mario Carneiro, 18-Jun-2015.) (Proof shortened by AV, 15-Jun-2019.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π Β· (2 Β· π₯)) mod π) & β’ π = (π₯ β (1...((π β 1) / 2)) β¦ ((((-1βπ ) Β· π ) mod π) / 2)) & β’ π = ((π Β· (2 Β· π¦)) mod π) & β’ π = (β€/nβ€βπ) & β’ πΊ = (mulGrpβπ) & β’ πΏ = (β€RHomβπ) β β’ (π β ((πβ((π β 1) / 2)) mod π) = ((-1βΞ£π₯ β (1...((π β 1) / 2))(ββ((π / π) Β· (2 Β· π₯)))) mod π)) | ||
Theorem | lgseisen 27325* | Eisenstein's lemma, an expression for (π /L π) when π, π are distinct odd primes. (Contributed by Mario Carneiro, 18-Jun-2015.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) β β’ (π β (π /L π) = (-1βΞ£π₯ β (1...((π β 1) / 2))(ββ((π / π) Β· (2 Β· π₯))))) | ||
Theorem | lgsquadlem1 27326* | Lemma for lgsquad 27329. Count the members of π with odd coordinates. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π β 1) / 2) & β’ π = ((π β 1) / 2) & β’ π = {β¨π₯, π¦β© β£ ((π₯ β (1...π) β§ π¦ β (1...π)) β§ (π¦ Β· π) < (π₯ Β· π))} β β’ (π β (-1βΞ£π’ β (((ββ(π / 2)) + 1)...π)(ββ((π / π) Β· (2 Β· π’)))) = (-1β(β―β{π§ β π β£ Β¬ 2 β₯ (1st βπ§)}))) | ||
Theorem | lgsquadlem2 27327* | Lemma for lgsquad 27329. Count the members of π with even coordinates, and combine with lgsquadlem1 27326 to get the total count of lattice points in π (up to parity). (Contributed by Mario Carneiro, 18-Jun-2015.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π β 1) / 2) & β’ π = ((π β 1) / 2) & β’ π = {β¨π₯, π¦β© β£ ((π₯ β (1...π) β§ π¦ β (1...π)) β§ (π¦ Β· π) < (π₯ Β· π))} β β’ (π β (π /L π) = (-1β(β―βπ))) | ||
Theorem | lgsquadlem3 27328* | Lemma for lgsquad 27329. (Contributed by Mario Carneiro, 18-Jun-2015.) |
β’ (π β π β (β β {2})) & β’ (π β π β (β β {2})) & β’ (π β π β π) & β’ π = ((π β 1) / 2) & β’ π = ((π β 1) / 2) & β’ π = {β¨π₯, π¦β© β£ ((π₯ β (1...π) β§ π¦ β (1...π)) β§ (π¦ Β· π) < (π₯ Β· π))} β β’ (π β ((π /L π) Β· (π /L π)) = (-1β(π Β· π))) | ||
Theorem | lgsquad 27329 | The Law of Quadratic Reciprocity, see also theorem 9.8 in [ApostolNT] p. 185. If π and π are distinct odd primes, then the product of the Legendre symbols (π /L π) and (π /L π) is the parity of ((π β 1) / 2) Β· ((π β 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.) |
β’ ((π β (β β {2}) β§ π β (β β {2}) β§ π β π) β ((π /L π) Β· (π /L π)) = (-1β(((π β 1) / 2) Β· ((π β 1) / 2)))) | ||
Theorem | lgsquad2lem1 27330 | Lemma for lgsquad2 27332. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ (π β π β β) & β’ (π β Β¬ 2 β₯ π) & β’ (π β π β β) & β’ (π β Β¬ 2 β₯ π) & β’ (π β (π gcd π) = 1) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β (π΄ Β· π΅) = π) & β’ (π β ((π΄ /L π) Β· (π /L π΄)) = (-1β(((π΄ β 1) / 2) Β· ((π β 1) / 2)))) & β’ (π β ((π΅ /L π) Β· (π /L π΅)) = (-1β(((π΅ β 1) / 2) Β· ((π β 1) / 2)))) β β’ (π β ((π /L π) Β· (π /L π)) = (-1β(((π β 1) / 2) Β· ((π β 1) / 2)))) | ||
Theorem | lgsquad2lem2 27331* | Lemma for lgsquad2 27332. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ (π β π β β) & β’ (π β Β¬ 2 β₯ π) & β’ (π β π β β) & β’ (π β Β¬ 2 β₯ π) & β’ (π β (π gcd π) = 1) & β’ ((π β§ (π β (β β {2}) β§ (π gcd π) = 1)) β ((π /L π) Β· (π /L π)) = (-1β(((π β 1) / 2) Β· ((π β 1) / 2)))) & β’ (π β βπ₯ β (1...π)((π₯ gcd (2 Β· π)) = 1 β ((π₯ /L π) Β· (π /L π₯)) = (-1β(((π₯ β 1) / 2) Β· ((π β 1) / 2))))) β β’ (π β ((π /L π) Β· (π /L π)) = (-1β(((π β 1) / 2) Β· ((π β 1) / 2)))) | ||
Theorem | lgsquad2 27332 | Extend lgsquad 27329 to coprime odd integers (the domain of the Jacobi symbol). (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ (π β π β β) & β’ (π β Β¬ 2 β₯ π) & β’ (π β π β β) & β’ (π β Β¬ 2 β₯ π) & β’ (π β (π gcd π) = 1) β β’ (π β ((π /L π) Β· (π /L π)) = (-1β(((π β 1) / 2) Β· ((π β 1) / 2)))) | ||
Theorem | lgsquad3 27333 | Extend lgsquad2 27332 to integers which share a factor. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ (((π β β β§ Β¬ 2 β₯ π) β§ (π β β β§ Β¬ 2 β₯ π)) β (π /L π) = ((-1β(((π β 1) / 2) Β· ((π β 1) / 2))) Β· (π /L π))) | ||
Theorem | m1lgs 27334 | The first supplement to the law of quadratic reciprocity. Negative one is a square mod an odd prime π iff πβ‘1 (mod 4). See first case of theorem 9.4 in [ApostolNT] p. 181. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ (π β (β β {2}) β ((-1 /L π) = 1 β (π mod 4) = 1)) | ||
Theorem | 2lgslem1a1 27335* | Lemma 1 for 2lgslem1a 27337. (Contributed by AV, 16-Jun-2021.) |
β’ ((π β β β§ Β¬ 2 β₯ π) β βπ β (1...((π β 1) / 2))(π Β· 2) = ((π Β· 2) mod π)) | ||
Theorem | 2lgslem1a2 27336 | Lemma 2 for 2lgslem1a 27337. (Contributed by AV, 18-Jun-2021.) |
β’ ((π β β€ β§ πΌ β β€) β ((ββ(π / 4)) < πΌ β (π / 2) < (πΌ Β· 2))) | ||
Theorem | 2lgslem1a 27337* | Lemma 1 for 2lgslem1 27340. (Contributed by AV, 18-Jun-2021.) |
β’ ((π β β β§ Β¬ 2 β₯ π) β {π₯ β β€ β£ βπ β (1...((π β 1) / 2))(π₯ = (π Β· 2) β§ (π / 2) < (π₯ mod π))} = {π₯ β β€ β£ βπ β (((ββ(π / 4)) + 1)...((π β 1) / 2))π₯ = (π Β· 2)}) | ||
Theorem | 2lgslem1b 27338* | Lemma 2 for 2lgslem1 27340. (Contributed by AV, 18-Jun-2021.) |
β’ πΌ = (π΄...π΅) & β’ πΉ = (π β πΌ β¦ (π Β· 2)) β β’ πΉ:πΌβ1-1-ontoβ{π₯ β β€ β£ βπ β πΌ π₯ = (π Β· 2)} | ||
Theorem | 2lgslem1c 27339 | Lemma 3 for 2lgslem1 27340. (Contributed by AV, 19-Jun-2021.) |
β’ ((π β β β§ Β¬ 2 β₯ π) β (ββ(π / 4)) β€ ((π β 1) / 2)) | ||
Theorem | 2lgslem1 27340* | Lemma 1 for 2lgs 27353. (Contributed by AV, 19-Jun-2021.) |
β’ ((π β β β§ Β¬ 2 β₯ π) β (β―β{π₯ β β€ β£ βπ β (1...((π β 1) / 2))(π₯ = (π Β· 2) β§ (π / 2) < (π₯ mod π))}) = (((π β 1) / 2) β (ββ(π / 4)))) | ||
Theorem | 2lgslem2 27341 | Lemma 2 for 2lgs 27353. (Contributed by AV, 20-Jun-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((π β β β§ Β¬ 2 β₯ π) β π β β€) | ||
Theorem | 2lgslem3a 27342 | Lemma for 2lgslem3a1 27346. (Contributed by AV, 14-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((πΎ β β0 β§ π = ((8 Β· πΎ) + 1)) β π = (2 Β· πΎ)) | ||
Theorem | 2lgslem3b 27343 | Lemma for 2lgslem3b1 27347. (Contributed by AV, 16-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((πΎ β β0 β§ π = ((8 Β· πΎ) + 3)) β π = ((2 Β· πΎ) + 1)) | ||
Theorem | 2lgslem3c 27344 | Lemma for 2lgslem3c1 27348. (Contributed by AV, 16-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((πΎ β β0 β§ π = ((8 Β· πΎ) + 5)) β π = ((2 Β· πΎ) + 1)) | ||
Theorem | 2lgslem3d 27345 | Lemma for 2lgslem3d1 27349. (Contributed by AV, 16-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((πΎ β β0 β§ π = ((8 Β· πΎ) + 7)) β π = ((2 Β· πΎ) + 2)) | ||
Theorem | 2lgslem3a1 27346 | Lemma 1 for 2lgslem3 27350. (Contributed by AV, 15-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((π β β β§ (π mod 8) = 1) β (π mod 2) = 0) | ||
Theorem | 2lgslem3b1 27347 | Lemma 2 for 2lgslem3 27350. (Contributed by AV, 16-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((π β β β§ (π mod 8) = 3) β (π mod 2) = 1) | ||
Theorem | 2lgslem3c1 27348 | Lemma 3 for 2lgslem3 27350. (Contributed by AV, 16-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((π β β β§ (π mod 8) = 5) β (π mod 2) = 1) | ||
Theorem | 2lgslem3d1 27349 | Lemma 4 for 2lgslem3 27350. (Contributed by AV, 15-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((π β β β§ (π mod 8) = 7) β (π mod 2) = 0) | ||
Theorem | 2lgslem3 27350 | Lemma 3 for 2lgs 27353. (Contributed by AV, 16-Jul-2021.) |
β’ π = (((π β 1) / 2) β (ββ(π / 4))) β β’ ((π β β β§ Β¬ 2 β₯ π) β (π mod 2) = if((π mod 8) β {1, 7}, 0, 1)) | ||
Theorem | 2lgs2 27351 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
β’ (2 /L 2) = 0 | ||
Theorem | 2lgslem4 27352 | Lemma 4 for 2lgs 27353: special case of 2lgs 27353 for π = 2. (Contributed by AV, 20-Jun-2021.) |
β’ ((2 /L 2) = 1 β (2 mod 8) β {1, 7}) | ||
Theorem | 2lgs 27353 | 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 π iff πβ‘Β±1 (mod 8), see first case of theorem 9.5 in [ApostolNT] p. 181. This theorem justifies our definition of (π /L 2) (lgs2 27260) to some degree, by demanding that reciprocity extend to the case π = 2. (Proposed by Mario Carneiro, 19-Jun-2015.) (Contributed by AV, 16-Jul-2021.) |
β’ (π β β β ((2 /L π) = 1 β (π mod 8) β {1, 7})) | ||
Theorem | 2lgsoddprmlem1 27354 | Lemma 1 for 2lgsoddprm 27362. (Contributed by AV, 19-Jul-2021.) |
β’ ((π΄ β β€ β§ π΅ β β€ β§ π = ((8 Β· π΄) + π΅)) β (((πβ2) β 1) / 8) = (((8 Β· (π΄β2)) + (2 Β· (π΄ Β· π΅))) + (((π΅β2) β 1) / 8))) | ||
Theorem | 2lgsoddprmlem2 27355 | Lemma 2 for 2lgsoddprm 27362. (Contributed by AV, 19-Jul-2021.) |
β’ ((π β β€ β§ Β¬ 2 β₯ π β§ π = (π mod 8)) β (2 β₯ (((πβ2) β 1) / 8) β 2 β₯ (((π β2) β 1) / 8))) | ||
Theorem | 2lgsoddprmlem3a 27356 | Lemma 1 for 2lgsoddprmlem3 27360. (Contributed by AV, 20-Jul-2021.) |
β’ (((1β2) β 1) / 8) = 0 | ||
Theorem | 2lgsoddprmlem3b 27357 | Lemma 2 for 2lgsoddprmlem3 27360. (Contributed by AV, 20-Jul-2021.) |
β’ (((3β2) β 1) / 8) = 1 | ||
Theorem | 2lgsoddprmlem3c 27358 | Lemma 3 for 2lgsoddprmlem3 27360. (Contributed by AV, 20-Jul-2021.) |
β’ (((5β2) β 1) / 8) = 3 | ||
Theorem | 2lgsoddprmlem3d 27359 | Lemma 4 for 2lgsoddprmlem3 27360. (Contributed by AV, 20-Jul-2021.) |
β’ (((7β2) β 1) / 8) = (2 Β· 3) | ||
Theorem | 2lgsoddprmlem3 27360 | Lemma 3 for 2lgsoddprm 27362. (Contributed by AV, 20-Jul-2021.) |
β’ ((π β β€ β§ Β¬ 2 β₯ π β§ π = (π mod 8)) β (2 β₯ (((π β2) β 1) / 8) β π β {1, 7})) | ||
Theorem | 2lgsoddprmlem4 27361 | Lemma 4 for 2lgsoddprm 27362. (Contributed by AV, 20-Jul-2021.) |
β’ ((π β β€ β§ Β¬ 2 β₯ π) β (2 β₯ (((πβ2) β 1) / 8) β (π mod 8) β {1, 7})) | ||
Theorem | 2lgsoddprm 27362 | 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 π) = -1^(((P^2)-1)/8) ). (Contributed by AV, 20-Jul-2021.) |
β’ (π β (β β {2}) β (2 /L π) = (-1β(((πβ2) β 1) / 8))) | ||
Theorem | 2sqlem1 27363* | Lemma for 2sq 27376. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) β β’ (π΄ β π β βπ₯ β β€[i] π΄ = ((absβπ₯)β2)) | ||
Theorem | 2sqlem2 27364* | Lemma for 2sq 27376. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) β β’ (π΄ β π β βπ₯ β β€ βπ¦ β β€ π΄ = ((π₯β2) + (π¦β2))) | ||
Theorem | mul2sq 27365 | 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.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) β β’ ((π΄ β π β§ π΅ β π) β (π΄ Β· π΅) β π) | ||
Theorem | 2sqlem3 27366 | Lemma for 2sqlem5 27368. (Contributed by Mario Carneiro, 20-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β πΆ β β€) & β’ (π β π· β β€) & β’ (π β (π Β· π) = ((π΄β2) + (π΅β2))) & β’ (π β π = ((πΆβ2) + (π·β2))) & β’ (π β π β₯ ((πΆ Β· π΅) + (π΄ Β· π·))) β β’ (π β π β π) | ||
Theorem | 2sqlem4 27367 | Lemma for 2sqlem5 27368. (Contributed by Mario Carneiro, 20-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β πΆ β β€) & β’ (π β π· β β€) & β’ (π β (π Β· π) = ((π΄β2) + (π΅β2))) & β’ (π β π = ((πΆβ2) + (π·β2))) β β’ (π β π β π) | ||
Theorem | 2sqlem5 27368 | Lemma for 2sq 27376. 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.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (π Β· π) β π) & β’ (π β π β π) β β’ (π β π β π) | ||
Theorem | 2sqlem6 27369* | Lemma for 2sq 27376. 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.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β βπ β β (π β₯ π΅ β π β π)) & β’ (π β (π΄ Β· π΅) β π) β β’ (π β π΄ β π) | ||
Theorem | 2sqlem7 27370* | Lemma for 2sq 27376. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ π = {π§ β£ βπ₯ β β€ βπ¦ β β€ ((π₯ gcd π¦) = 1 β§ π§ = ((π₯β2) + (π¦β2)))} β β’ π β (π β© β) | ||
Theorem | 2sqlem8a 27371* | Lemma for 2sqlem8 27372. (Contributed by Mario Carneiro, 4-Jun-2016.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ π = {π§ β£ βπ₯ β β€ βπ¦ β β€ ((π₯ gcd π¦) = 1 β§ π§ = ((π₯β2) + (π¦β2)))} & β’ (π β βπ β (1...(π β 1))βπ β π (π β₯ π β π β π)) & β’ (π β π β₯ π) & β’ (π β π β β) & β’ (π β π β (β€β₯β2)) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β (π΄ gcd π΅) = 1) & β’ (π β π = ((π΄β2) + (π΅β2))) & β’ πΆ = (((π΄ + (π / 2)) mod π) β (π / 2)) & β’ π· = (((π΅ + (π / 2)) mod π) β (π / 2)) β β’ (π β (πΆ gcd π·) β β) | ||
Theorem | 2sqlem8 27372* | Lemma for 2sq 27376. (Contributed by Mario Carneiro, 20-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ π = {π§ β£ βπ₯ β β€ βπ¦ β β€ ((π₯ gcd π¦) = 1 β§ π§ = ((π₯β2) + (π¦β2)))} & β’ (π β βπ β (1...(π β 1))βπ β π (π β₯ π β π β π)) & β’ (π β π β₯ π) & β’ (π β π β β) & β’ (π β π β (β€β₯β2)) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β (π΄ gcd π΅) = 1) & β’ (π β π = ((π΄β2) + (π΅β2))) & β’ πΆ = (((π΄ + (π / 2)) mod π) β (π / 2)) & β’ π· = (((π΅ + (π / 2)) mod π) β (π / 2)) & β’ πΈ = (πΆ / (πΆ gcd π·)) & β’ πΉ = (π· / (πΆ gcd π·)) β β’ (π β π β π) | ||
Theorem | 2sqlem9 27373* | Lemma for 2sq 27376. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ π = {π§ β£ βπ₯ β β€ βπ¦ β β€ ((π₯ gcd π¦) = 1 β§ π§ = ((π₯β2) + (π¦β2)))} & β’ (π β βπ β (1...(π β 1))βπ β π (π β₯ π β π β π)) & β’ (π β π β₯ π) & β’ (π β π β β) & β’ (π β π β π) β β’ (π β π β π) | ||
Theorem | 2sqlem10 27374* | Lemma for 2sq 27376. 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.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ π = {π§ β£ βπ₯ β β€ βπ¦ β β€ ((π₯ gcd π¦) = 1 β§ π§ = ((π₯β2) + (π¦β2)))} β β’ ((π΄ β π β§ π΅ β β β§ π΅ β₯ π΄) β π΅ β π) | ||
Theorem | 2sqlem11 27375* | Lemma for 2sq 27376. (Contributed by Mario Carneiro, 19-Jun-2015.) |
β’ π = ran (π€ β β€[i] β¦ ((absβπ€)β2)) & β’ π = {π§ β£ βπ₯ β β€ βπ¦ β β€ ((π₯ gcd π¦) = 1 β§ π§ = ((π₯β2) + (π¦β2)))} β β’ ((π β β β§ (π mod 4) = 1) β π β π) | ||
Theorem | 2sq 27376* | All primes of the form 4π + 1 are sums of two squares. This is Metamath 100 proof #20. (Contributed by Mario Carneiro, 20-Jun-2015.) |
β’ ((π β β β§ (π mod 4) = 1) β βπ₯ β β€ βπ¦ β β€ π = ((π₯β2) + (π¦β2))) | ||
Theorem | 2sqblem 27377 | Lemma for 2sqb 27378. (Contributed by Mario Carneiro, 20-Jun-2015.) |
β’ (π β (π β β β§ π β 2)) & β’ (π β (π β β€ β§ π β β€)) & β’ (π β π = ((πβ2) + (πβ2))) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β (π gcd π) = ((π Β· π΄) + (π Β· π΅))) β β’ (π β (π mod 4) = 1) | ||
Theorem | 2sqb 27378* | The converse to 2sq 27376. (Contributed by Mario Carneiro, 20-Jun-2015.) |
β’ (π β β β (βπ₯ β β€ βπ¦ β β€ π = ((π₯β2) + (π¦β2)) β (π = 2 β¨ (π mod 4) = 1))) | ||
Theorem | 2sq2 27379 | 2 is the sum of squares of two nonnegative integers iff the two integers are 1. (Contributed by AV, 19-Jun-2023.) |
β’ ((π΄ β β0 β§ π΅ β β0) β (((π΄β2) + (π΅β2)) = 2 β (π΄ = 1 β§ π΅ = 1))) | ||
Theorem | 2sqn0 27380 | If the sum of two squares is prime, none of the original number is zero. (Contributed by Thierry Arnoux, 4-Feb-2020.) |
β’ (π β π β β) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β ((π΄β2) + (π΅β2)) = π) β β’ (π β π΄ β 0) | ||
Theorem | 2sqcoprm 27381 | If the sum of two squares is prime, the two original numbers are coprime. (Contributed by Thierry Arnoux, 2-Feb-2020.) |
β’ (π β π β β) & β’ (π β π΄ β β€) & β’ (π β π΅ β β€) & β’ (π β ((π΄β2) + (π΅β2)) = π) β β’ (π β (π΄ gcd π΅) = 1) | ||
Theorem | 2sqmod 27382 | Given two decompositions of a prime as a sum of two squares, show that they are equal. (Contributed by Thierry Arnoux, 2-Feb-2020.) |
β’ (π β π β β) & β’ (π β π΄ β β0) & β’ (π β π΅ β β0) & β’ (π β πΆ β β0) & β’ (π β π· β β0) & β’ (π β π΄ β€ π΅) & β’ (π β πΆ β€ π·) & β’ (π β ((π΄β2) + (π΅β2)) = π) & β’ (π β ((πΆβ2) + (π·β2)) = π) β β’ (π β (π΄ = πΆ β§ π΅ = π·)) | ||
Theorem | 2sqmo 27383* | There exists at most one decomposition of a prime as a sum of two squares. See 2sqb 27378 for the existence of such a decomposition. (Contributed by Thierry Arnoux, 2-Feb-2020.) |
β’ (π β β β β*π β β0 βπ β β0 (π β€ π β§ ((πβ2) + (πβ2)) = π)) | ||
Theorem | 2sqnn0 27384* | All primes of the form 4π + 1 are sums of squares of two nonnegative integers. (Contributed by AV, 3-Jun-2023.) |
β’ ((π β β β§ (π mod 4) = 1) β βπ₯ β β0 βπ¦ β β0 π = ((π₯β2) + (π¦β2))) | ||
Theorem | 2sqnn 27385* | All primes of the form 4π + 1 are sums of squares of two positive integers. (Contributed by AV, 11-Jun-2023.) |
β’ ((π β β β§ (π mod 4) = 1) β βπ₯ β β βπ¦ β β π = ((π₯β2) + (π¦β2))) | ||
Theorem | addsq2reu 27386* |
For each complex number πΆ, there exists a unique complex
number
π added to the square of a unique
another complex number π
resulting in the given complex number πΆ. The unique complex number
π is πΆ, and the unique another complex
number π is 0.
Remark: This, together with addsqnreup 27389, is an example showing that the pattern β!π β π΄β!π β π΅π does not necessarily mean "There are unique sets π and π fulfilling π). See also comments for df-eu 2557 and 2eu4 2643. For more details see comment for addsqnreup 27389. (Contributed by AV, 21-Jun-2023.) |
β’ (πΆ β β β β!π β β β!π β β (π + (πβ2)) = πΆ) | ||
Theorem | addsqn2reu 27387* |
For each complex number πΆ, there does not exist a unique
complex
number π, squared and added to a unique
another complex number
π resulting in the given complex number
πΆ.
Actually, for each
complex number π, π = (πΆ β (πβ2)) is unique.
Remark: This, together with addsq2reu 27386, 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.) |
β’ (πΆ β β β Β¬ β!π β β β!π β β (π + (πβ2)) = πΆ) | ||
Theorem | addsqrexnreu 27388* |
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 27386, 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 27386). For more details see comment for addsqnreup 27389. (Contributed by AV, 20-Jun-2023.) |
β’ (πΆ β β β βπ β β Β¬ β!π β β (π + (πβ2)) = πΆ) | ||
Theorem | addsqnreup 27389* |
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 27386, is a real life example (about a numerical property) showing that the pattern β!π β π΄β!π β π΅π does not necessarily mean "There are unique sets π and π fulfilling π"). 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 27386). 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 27388). For example, β¨1, (ββ(πΆ β 1))β© and β¨1, -(ββ(πΆ β 1))β© are two different decompositions of πΆ (if πΆ β 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 π and π so that .." (more formally β!π β (π΄ Γ π΅)π with π = β¨π, πβ©), or by showing (β!π₯ β π΄βπ¦ β π΅π β§ β!π¦ β π΅βπ₯ β π΄π) (see 2reu4 4523 resp. 2eu4 2643). These two representations are equivalent (see opreu2reurex 6294). An analogon of this theorem using the latter variant is given in addsqn2reurex2 27391. In some cases, however, the variant with (ordered!) pairs may be possible only for ordered sets (like β or β) and claiming that the first component is less than or equal to the second component (see, for example, 2sqreunnltb 27407 and 2sqreuopb 27414). Alternatively, (proper) unordered pairs can be used: β!πππ« π΄((β―βπ) = 2 β§ π), or, using the definition of proper pairs: β!π β (Pairsproperβπ΄)π (see, for example, inlinecirc02preu 47969). (Contributed by AV, 21-Jun-2023.) |
β’ (πΆ β β β Β¬ β!π β (β Γ β)((1st βπ) + ((2nd βπ)β2)) = πΆ) | ||
Theorem | addsq2nreurex 27390* | For each complex number πΆ, there is no unique complex number π added to the square of another complex number π resulting in the given complex number πΆ. (Contributed by AV, 2-Jul-2023.) |
β’ (πΆ β β β Β¬ β!π β β βπ β β (π + (πβ2)) = πΆ) | ||
Theorem | addsqn2reurex2 27391* |
For each complex number πΆ, there does not uniquely exist two
complex numbers π and π, with π squared
and added to π
resulting in the given complex number πΆ.
Remark: This, together with addsq2reu 27386, is an example showing that the pattern β!π β π΄β!π β π΅π does not necessarily mean "There are unique sets π and π fulfilling π), as it is the case with the pattern (β!π β π΄βπ β π΅π β§ β!π β π΅βπ β π΄π. See also comments for df-eu 2557 and 2eu4 2643. (Contributed by AV, 2-Jul-2023.) |
β’ (πΆ β β β Β¬ (β!π β β βπ β β (π + (πβ2)) = πΆ β§ β!π β β βπ β β (π + (πβ2)) = πΆ)) | ||
Theorem | 2sqreulem1 27392* | Lemma 1 for 2sqreu 27402. (Contributed by AV, 4-Jun-2023.) |
β’ ((π β β β§ (π mod 4) = 1) β β!π β β0 β!π β β0 (π β€ π β§ ((πβ2) + (πβ2)) = π)) | ||
Theorem | 2sqreultlem 27393* | Lemma for 2sqreult 27404. (Contributed by AV, 8-Jun-2023.) (Proposed by GL, 8-Jun-2023.) |
β’ ((π β β β§ (π mod 4) = 1) β β!π β β0 β!π β β0 (π < π β§ ((πβ2) + (πβ2)) = π)) | ||
Theorem | 2sqreultblem 27394* | Lemma for 2sqreultb 27405. (Contributed by AV, 10-Jun-2023.) The prime needs not be odd, as observed by WL. (Revised by AV, 18-Jun-2023.) |
β’ (π β β β ((π mod 4) = 1 β β!π β β0 β!π β β0 (π < π β§ ((πβ2) + (πβ2)) = π))) | ||
Theorem | 2sqreunnlem1 27395* | Lemma 1 for 2sqreunn 27403. (Contributed by AV, 11-Jun-2023.) |
β’ ((π β β β§ (π mod 4) = 1) β β!π β β β!π β β (π β€ π β§ ((πβ2) + (πβ2)) = π)) | ||
Theorem | 2sqreunnltlem 27396* | Lemma for 2sqreunnlt 27406. (Contributed by AV, 4-Jun-2023.) Specialization to different integers, proposed by GL. (Revised by AV, 11-Jun-2023.) |
β’ ((π β β β§ (π mod 4) = 1) β β!π β β β!π β β (π < π β§ ((πβ2) + (πβ2)) = π)) | ||
Theorem | 2sqreunnltblem 27397* | Lemma for 2sqreunnltb 27407. (Contributed by AV, 11-Jun-2023.) The prime needs not be odd, as observed by WL. (Revised by AV, 18-Jun-2023.) |
β’ (π β β β ((π mod 4) = 1 β β!π β β β!π β β (π < π β§ ((πβ2) + (πβ2)) = π))) | ||
Theorem | 2sqreulem2 27398 | Lemma 2 for 2sqreu 27402 etc. (Contributed by AV, 25-Jun-2023.) |
β’ ((π΄ β β0 β§ π΅ β β0 β§ πΆ β β0) β (((π΄β2) + (π΅β2)) = ((π΄β2) + (πΆβ2)) β π΅ = πΆ)) | ||
Theorem | 2sqreulem3 27399 | Lemma 3 for 2sqreu 27402 etc. (Contributed by AV, 25-Jun-2023.) |
β’ ((π΄ β β0 β§ (π΅ β β0 β§ πΆ β β0)) β (((π β§ ((π΄β2) + (π΅β2)) = π) β§ (π β§ ((π΄β2) + (πΆβ2)) = π)) β π΅ = πΆ)) | ||
Theorem | 2sqreulem4 27400* | Lemma 4 for 2sqreu 27402 et. (Contributed by AV, 25-Jun-2023.) |
β’ (π β (π β§ ((πβ2) + (πβ2)) = π)) β β’ βπ β β0 β*π β β0 π |
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