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Type | Label | Description |
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Statement | ||
Theorem | lgscl2 26801* | The Legendre symbol is an integer with absolute value less than or equal to 1. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ π = {π₯ β β€ β£ (absβπ₯) β€ 1} β β’ ((π΄ β β€ β§ π β β€) β (π΄ /L π) β π) | ||
Theorem | lgs0 26802 | The Legendre symbol when the second argument is zero. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (π΄ β β€ β (π΄ /L 0) = if((π΄β2) = 1, 1, 0)) | ||
Theorem | lgscl 26803 | The Legendre symbol is an integer. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β€) β (π΄ /L π) β β€) | ||
Theorem | lgsle1 26804 | The Legendre symbol has absolute value less than or equal to 1. Together with lgscl 26803 this implies that it takes values in {-1, 0, 1}. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β€) β (absβ(π΄ /L π)) β€ 1) | ||
Theorem | lgsval2 26805 | The Legendre symbol at a prime (this is the traditional domain of the Legendre symbol, except for the addition of prime 2). (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β) β (π΄ /L π) = if(π = 2, if(2 β₯ π΄, 0, if((π΄ mod 8) β {1, 7}, 1, -1)), ((((π΄β((π β 1) / 2)) + 1) mod π) β 1))) | ||
Theorem | lgs2 26806 | The Legendre symbol at 2. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (π΄ β β€ β (π΄ /L 2) = if(2 β₯ π΄, 0, if((π΄ mod 8) β {1, 7}, 1, -1))) | ||
Theorem | lgsval3 26807 | The Legendre symbol at an odd prime (this is the traditional domain of the Legendre symbol). (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β (β β {2})) β (π΄ /L π) = ((((π΄β((π β 1) / 2)) + 1) mod π) β 1)) | ||
Theorem | lgsvalmod 26808 | The Legendre symbol is equivalent to πβ((π β 1) / 2), mod π. This theorem is also called "Euler's criterion", see theorem 9.2 in [ApostolNT] p. 180, or a representation of Euler's criterion using the Legendre symbol, see also lgsqr 26843. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β (β β {2})) β ((π΄ /L π) mod π) = ((π΄β((π β 1) / 2)) mod π)) | ||
Theorem | lgsval4 26809* | Restate lgsval 26793 for nonzero π, where the function πΉ has been abbreviated into a self-referential expression taking the value of /L on the primes as given. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ πΉ = (π β β β¦ if(π β β, ((π΄ /L π)β(π pCnt π)), 1)) β β’ ((π΄ β β€ β§ π β β€ β§ π β 0) β (π΄ /L π) = (if((π < 0 β§ π΄ < 0), -1, 1) Β· (seq1( Β· , πΉ)β(absβπ)))) | ||
Theorem | lgsfcl3 26810* | Closure of the function πΉ which defines the Legendre symbol at the primes. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ πΉ = (π β β β¦ if(π β β, ((π΄ /L π)β(π pCnt π)), 1)) β β’ ((π΄ β β€ β§ π β β€ β§ π β 0) β πΉ:ββΆβ€) | ||
Theorem | lgsval4a 26811* | Same as lgsval4 26809 for positive π. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ πΉ = (π β β β¦ if(π β β, ((π΄ /L π)β(π pCnt π)), 1)) β β’ ((π΄ β β€ β§ π β β) β (π΄ /L π) = (seq1( Β· , πΉ)βπ)) | ||
Theorem | lgscl1 26812 | The value of the Legendre symbol is either -1 or 0 or 1. (Contributed by AV, 13-Jul-2021.) |
β’ ((π΄ β β€ β§ π β β€) β (π΄ /L π) β {-1, 0, 1}) | ||
Theorem | lgsneg 26813 | The Legendre symbol is either even or odd under negation with respect to the second parameter according to the sign of the first. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β€ β§ π β 0) β (π΄ /L -π) = (if(π΄ < 0, -1, 1) Β· (π΄ /L π))) | ||
Theorem | lgsneg1 26814 | The Legendre symbol for nonnegative first parameter is unchanged by negation of the second. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β0 β§ π β β€) β (π΄ /L -π) = (π΄ /L π)) | ||
Theorem | lgsmod 26815 | The Legendre (Jacobi) symbol is preserved under reduction mod π when π is odd. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β β§ Β¬ 2 β₯ π) β ((π΄ mod π) /L π) = (π΄ /L π)) | ||
Theorem | lgsdilem 26816 | Lemma for lgsdi 26826 and lgsdir 26824: the sign part of the Legendre symbol is multiplicative. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (((π΄ β β€ β§ π΅ β β€ β§ π β β€) β§ (π΄ β 0 β§ π΅ β 0)) β if((π < 0 β§ (π΄ Β· π΅) < 0), -1, 1) = (if((π < 0 β§ π΄ < 0), -1, 1) Β· if((π < 0 β§ π΅ < 0), -1, 1))) | ||
Theorem | lgsdir2lem1 26817 | Lemma for lgsdir2 26822. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (((1 mod 8) = 1 β§ (-1 mod 8) = 7) β§ ((3 mod 8) = 3 β§ (-3 mod 8) = 5)) | ||
Theorem | lgsdir2lem2 26818 | Lemma for lgsdir2 26822. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (πΎ β β€ β§ 2 β₯ (πΎ + 1) β§ ((π΄ β β€ β§ Β¬ 2 β₯ π΄) β ((π΄ mod 8) β (0...πΎ) β (π΄ mod 8) β π))) & β’ π = (πΎ + 1) & β’ π = (π + 1) & β’ π β π β β’ (π β β€ β§ 2 β₯ (π + 1) β§ ((π΄ β β€ β§ Β¬ 2 β₯ π΄) β ((π΄ mod 8) β (0...π) β (π΄ mod 8) β π))) | ||
Theorem | lgsdir2lem3 26819 | Lemma for lgsdir2 26822. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ Β¬ 2 β₯ π΄) β (π΄ mod 8) β ({1, 7} βͺ {3, 5})) | ||
Theorem | lgsdir2lem4 26820 | Lemma for lgsdir2 26822. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (((π΄ β β€ β§ π΅ β β€) β§ (π΄ mod 8) β {1, 7}) β (((π΄ Β· π΅) mod 8) β {1, 7} β (π΅ mod 8) β {1, 7})) | ||
Theorem | lgsdir2lem5 26821 | Lemma for lgsdir2 26822. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (((π΄ β β€ β§ π΅ β β€) β§ ((π΄ mod 8) β {3, 5} β§ (π΅ mod 8) β {3, 5})) β ((π΄ Β· π΅) mod 8) β {1, 7}) | ||
Theorem | lgsdir2 26822 | The Legendre symbol is completely multiplicative at 2. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ ((π΄ β β€ β§ π΅ β β€) β ((π΄ Β· π΅) /L 2) = ((π΄ /L 2) Β· (π΅ /L 2))) | ||
Theorem | lgsdirprm 26823 | The Legendre symbol is completely multiplicative at the primes. See theorem 9.3 in [ApostolNT] p. 180. (Contributed by Mario Carneiro, 4-Feb-2015.) (Proof shortened by AV, 18-Mar-2022.) |
β’ ((π΄ β β€ β§ π΅ β β€ β§ π β β) β ((π΄ Β· π΅) /L π) = ((π΄ /L π) Β· (π΅ /L π))) | ||
Theorem | lgsdir 26824 | The Legendre symbol is completely multiplicative in its left argument. Generalization of theorem 9.9(a) in [ApostolNT] p. 188 (which assumes that π΄ and π΅ are odd positive integers). Together with lgsqr 26843 this implies that the product of two quadratic residues or nonresidues is a residue, and the product of a residue and a nonresidue is a nonresidue. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (((π΄ β β€ β§ π΅ β β€ β§ π β β€) β§ (π΄ β 0 β§ π΅ β 0)) β ((π΄ Β· π΅) /L π) = ((π΄ /L π) Β· (π΅ /L π))) | ||
Theorem | lgsdilem2 26825* | Lemma for lgsdi 26826. (Contributed by Mario Carneiro, 4-Feb-2015.) |
β’ (π β π΄ β β€) & β’ (π β π β β€) & β’ (π β π β β€) & β’ (π β π β 0) & β’ (π β π β 0) & β’ πΉ = (π β β β¦ if(π β β, ((π΄ /L π)β(π pCnt π)), 1)) β β’ (π β (seq1( Β· , πΉ)β(absβπ)) = (seq1( Β· , πΉ)β(absβ(π Β· π)))) | ||
Theorem | lgsdi 26826 | The Legendre symbol is completely multiplicative in its right argument. Generalization of theorem 9.9(b) in [ApostolNT] p. 188 (which assumes that π and π are odd positive integers). (Contributed by Mario Carneiro, 5-Feb-2015.) |
β’ (((π΄ β β€ β§ π β β€ β§ π β β€) β§ (π β 0 β§ π β 0)) β (π΄ /L (π Β· π)) = ((π΄ /L π) Β· (π΄ /L π))) | ||
Theorem | lgsne0 26827 | The Legendre symbol is nonzero (and hence equal to 1 or -1) precisely when the arguments are coprime. (Contributed by Mario Carneiro, 5-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β€) β ((π΄ /L π) β 0 β (π΄ gcd π) = 1)) | ||
Theorem | lgsabs1 26828 | The Legendre symbol is nonzero (and hence equal to 1 or -1) precisely when the arguments are coprime. (Contributed by Mario Carneiro, 5-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β€) β ((absβ(π΄ /L π)) = 1 β (π΄ gcd π) = 1)) | ||
Theorem | lgssq 26829 | The Legendre symbol at a square is equal to 1. Together with lgsmod 26815 this implies that the Legendre symbol takes value 1 at every quadratic residue. (Contributed by Mario Carneiro, 5-Feb-2015.) (Revised by AV, 20-Jul-2021.) |
β’ (((π΄ β β€ β§ π΄ β 0) β§ π β β€ β§ (π΄ gcd π) = 1) β ((π΄β2) /L π) = 1) | ||
Theorem | lgssq2 26830 | The Legendre symbol at a square is equal to 1. (Contributed by Mario Carneiro, 5-Feb-2015.) |
β’ ((π΄ β β€ β§ π β β β§ (π΄ gcd π) = 1) β (π΄ /L (πβ2)) = 1) | ||
Theorem | lgsprme0 26831 | The Legendre symbol at any prime (even at 2) is 0 iff the prime does not divide the first argument. See definition in [ApostolNT] p. 179. (Contributed by AV, 20-Jul-2021.) |
β’ ((π΄ β β€ β§ π β β) β ((π΄ /L π) = 0 β (π΄ mod π) = 0)) | ||
Theorem | 1lgs 26832 | The Legendre symbol at 1. See example 1 in [ApostolNT] p. 180. (Contributed by Mario Carneiro, 28-Apr-2016.) |
β’ (π β β€ β (1 /L π) = 1) | ||
Theorem | lgs1 26833 | The Legendre symbol at 1. See definition in [ApostolNT] p. 188. (Contributed by Mario Carneiro, 28-Apr-2016.) |
β’ (π΄ β β€ β (π΄ /L 1) = 1) | ||
Theorem | lgsmodeq 26834 | The Legendre (Jacobi) symbol is preserved under reduction mod π when π is odd. Theorem 9.9(c) in [ApostolNT] p. 188. (Contributed by AV, 20-Jul-2021.) |
β’ ((π΄ β β€ β§ π΅ β β€ β§ (π β β β§ Β¬ 2 β₯ π)) β ((π΄ mod π) = (π΅ mod π) β (π΄ /L π) = (π΅ /L π))) | ||
Theorem | lgsmulsqcoprm 26835 | The Legendre (Jacobi) symbol is preserved under multiplication with a square of an integer coprime to the second argument. Theorem 9.9(d) in [ApostolNT] p. 188. (Contributed by AV, 20-Jul-2021.) |
β’ (((π΄ β β€ β§ π΄ β 0) β§ (π΅ β β€ β§ π΅ β 0) β§ (π β β€ β§ (π΄ gcd π) = 1)) β (((π΄β2) Β· π΅) /L π) = (π΅ /L π)) | ||
Theorem | lgsdirnn0 26836 | Variation on lgsdir 26824 valid for all π΄, π΅ but only for positive π. (The exact location of the failure of this law is for π΄ = 0, π΅ < 0, π = -1 in which case (0 /L -1) = 1 but (π΅ /L -1) = -1.) (Contributed by Mario Carneiro, 28-Apr-2016.) |
β’ ((π΄ β β€ β§ π΅ β β€ β§ π β β0) β ((π΄ Β· π΅) /L π) = ((π΄ /L π) Β· (π΅ /L π))) | ||
Theorem | lgsdinn0 26837 | Variation on lgsdi 26826 valid for all π, π but only for positive π΄. (The exact location of the failure of this law is for π΄ = -1, π = 0, and some π in which case (-1 /L 0) = 1 but (-1 /L π) = -1 when -1 is not a quadratic residue mod π.) (Contributed by Mario Carneiro, 28-Apr-2016.) |
β’ ((π΄ β β0 β§ π β β€ β§ π β β€) β (π΄ /L (π Β· π)) = ((π΄ /L π) Β· (π΄ /L π))) | ||
Theorem | lgsqrlem1 26838 | Lemma for lgsqr 26843. (Contributed by Mario Carneiro, 15-Jun-2015.) |
β’ π = (β€/nβ€βπ) & β’ π = (Poly1βπ) & β’ π΅ = (Baseβπ) & β’ π· = ( deg1 βπ) & β’ π = (eval1βπ) & β’ β = (.gβ(mulGrpβπ)) & β’ π = (var1βπ) & β’ β = (-gβπ) & β’ 1 = (1rβπ) & β’ π = ((((π β 1) / 2) β π) β 1 ) & β’ πΏ = (β€RHomβπ) & β’ (π β π β (β β {2})) & β’ (π β π΄ β β€) & β’ (π β ((π΄β((π β 1) / 2)) mod π) = (1 mod π)) β β’ (π β ((πβπ)β(πΏβπ΄)) = (0gβπ)) | ||
Theorem | lgsqrlem2 26839* | Lemma for lgsqr 26843. (Contributed by Mario Carneiro, 15-Jun-2015.) |
β’ π = (β€/nβ€βπ) & β’ π = (Poly1βπ) & β’ π΅ = (Baseβπ) & β’ π· = ( deg1 βπ) & β’ π = (eval1βπ) & β’ β = (.gβ(mulGrpβπ)) & β’ π = (var1βπ) & β’ β = (-gβπ) & β’ 1 = (1rβπ) & β’ π = ((((π β 1) / 2) β π) β 1 ) & β’ πΏ = (β€RHomβπ) & β’ (π β π β (β β {2})) & β’ πΊ = (π¦ β (1...((π β 1) / 2)) β¦ (πΏβ(π¦β2))) β β’ (π β πΊ:(1...((π β 1) / 2))β1-1β(β‘(πβπ) β {(0gβπ)})) | ||
Theorem | lgsqrlem3 26840* | Lemma for lgsqr 26843. (Contributed by Mario Carneiro, 15-Jun-2015.) |
β’ π = (β€/nβ€βπ) & β’ π = (Poly1βπ) & β’ π΅ = (Baseβπ) & β’ π· = ( deg1 βπ) & β’ π = (eval1βπ) & β’ β = (.gβ(mulGrpβπ)) & β’ π = (var1βπ) & β’ β = (-gβπ) & β’ 1 = (1rβπ) & β’ π = ((((π β 1) / 2) β π) β 1 ) & β’ πΏ = (β€RHomβπ) & β’ (π β π β (β β {2})) & β’ πΊ = (π¦ β (1...((π β 1) / 2)) β¦ (πΏβ(π¦β2))) & β’ (π β π΄ β β€) & β’ (π β (π΄ /L π) = 1) β β’ (π β (πΏβπ΄) β (β‘(πβπ) β {(0gβπ)})) | ||
Theorem | lgsqrlem4 26841* | Lemma for lgsqr 26843. (Contributed by Mario Carneiro, 15-Jun-2015.) |
β’ π = (β€/nβ€βπ) & β’ π = (Poly1βπ) & β’ π΅ = (Baseβπ) & β’ π· = ( deg1 βπ) & β’ π = (eval1βπ) & β’ β = (.gβ(mulGrpβπ)) & β’ π = (var1βπ) & β’ β = (-gβπ) & β’ 1 = (1rβπ) & β’ π = ((((π β 1) / 2) β π) β 1 ) & β’ πΏ = (β€RHomβπ) & β’ (π β π β (β β {2})) & β’ πΊ = (π¦ β (1...((π β 1) / 2)) β¦ (πΏβ(π¦β2))) & β’ (π β π΄ β β€) & β’ (π β (π΄ /L π) = 1) β β’ (π β βπ₯ β β€ π β₯ ((π₯β2) β π΄)) | ||
Theorem | lgsqrlem5 26842* | Lemma for lgsqr 26843. (Contributed by Mario Carneiro, 15-Jun-2015.) |
β’ ((π΄ β β€ β§ π β (β β {2}) β§ (π΄ /L π) = 1) β βπ₯ β β€ π β₯ ((π₯β2) β π΄)) | ||
Theorem | lgsqr 26843* | The Legendre symbol for odd primes is 1 iff the number is not a multiple of the prime (in which case it is 0, see lgsne0 26827) and the number is a quadratic residue mod π (it is -1 for nonresidues by the process of elimination from lgsabs1 26828). Given our definition of the Legendre symbol, this theorem is equivalent to Euler's criterion. (Contributed by Mario Carneiro, 15-Jun-2015.) |
β’ ((π΄ β β€ β§ π β (β β {2})) β ((π΄ /L π) = 1 β (Β¬ π β₯ π΄ β§ βπ₯ β β€ π β₯ ((π₯β2) β π΄)))) | ||
Theorem | lgsqrmod 26844* | If the Legendre symbol of an integer for an odd prime is 1, then the number is a quadratic residue mod π. (Contributed by AV, 20-Aug-2021.) |
β’ ((π΄ β β€ β§ π β (β β {2})) β ((π΄ /L π) = 1 β βπ₯ β β€ ((π₯β2) mod π) = (π΄ mod π))) | ||
Theorem | lgsqrmodndvds 26845* | If the Legendre symbol of an integer π΄ for an odd prime is 1, then the number is a quadratic residue mod π with a solution π₯ of the congruence (π₯β2)β‘π΄ (mod π) which is not divisible by the prime. (Contributed by AV, 20-Aug-2021.) (Proof shortened by AV, 18-Mar-2022.) |
β’ ((π΄ β β€ β§ π β (β β {2})) β ((π΄ /L π) = 1 β βπ₯ β β€ (((π₯β2) mod π) = (π΄ mod π) β§ Β¬ π β₯ π₯))) | ||
Theorem | lgsdchrval 26846* | The Legendre symbol function π(π) = (π /L π), where π is an odd positive number, is a 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 β€) β (πβ(πΏβπ΄)) = (π΄ /L π)) | ||
Theorem | lgsdchr 26847* | 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 26866. The general case is still to prove. | ||
Theorem | gausslemma2dlem0a 26848 | Auxiliary lemma 1 for gausslemma2d 26866. (Contributed by AV, 9-Jul-2021.) |
β’ (π β π β (β β {2})) β β’ (π β π β β) | ||
Theorem | gausslemma2dlem0b 26849 | Auxiliary lemma 2 for gausslemma2d 26866. (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 26850 | Auxiliary lemma 3 for gausslemma2d 26866. (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 26851 | Auxiliary lemma 4 for gausslemma2d 26866. (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 26852 | Auxiliary lemma 5 for gausslemma2d 26866. (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 26853 | Auxiliary lemma 6 for gausslemma2d 26866. (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 26854 | Auxiliary lemma 7 for gausslemma2d 26866. (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 26855 | Auxiliary lemma 8 for gausslemma2d 26866. (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 26856 | Auxiliary lemma 9 for gausslemma2d 26866. (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 26857* | Lemma for gausslemma2dlem1 26858. (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 26858* | Lemma 1 for gausslemma2d 26866. (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 26859* | Lemma 2 for gausslemma2d 26866. (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 26860* | Lemma 3 for gausslemma2d 26866. (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 26861* | Lemma 4 for gausslemma2d 26866. (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 26862* | Lemma for gausslemma2dlem5 26863. (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 26863* | Lemma 5 for gausslemma2d 26866. (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 26864* | Lemma 6 for gausslemma2d 26866. (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 26865* | Lemma 7 for gausslemma2d 26866. (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 26866* | 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 26867* | Lemma for lgseisen 26871. 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 26868* | Lemma for lgseisen 26871. 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 26869* | Lemma for lgseisen 26871. (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 26870* | Lemma for lgseisen 26871. 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 26871* | 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 26872* | Lemma for lgsquad 26875. 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 26873* | Lemma for lgsquad 26875. Count the members of π with even coordinates, and combine with lgsquadlem1 26872 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 26874* | Lemma for lgsquad 26875. (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 26875 | 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 26876 | Lemma for lgsquad2 26878. (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 26877* | Lemma for lgsquad2 26878. (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 26878 | Extend lgsquad 26875 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 26879 | Extend lgsquad2 26878 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 26880 | 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 26881* | Lemma 1 for 2lgslem1a 26883. (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 26882 | Lemma 2 for 2lgslem1a 26883. (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 26883* | Lemma 1 for 2lgslem1 26886. (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 26884* | Lemma 2 for 2lgslem1 26886. (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 26885 | Lemma 3 for 2lgslem1 26886. (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 26886* | Lemma 1 for 2lgs 26899. (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 26887 | Lemma 2 for 2lgs 26899. (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 26888 | Lemma for 2lgslem3a1 26892. (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 26889 | Lemma for 2lgslem3b1 26893. (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 26890 | Lemma for 2lgslem3c1 26894. (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 26891 | Lemma for 2lgslem3d1 26895. (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 26892 | Lemma 1 for 2lgslem3 26896. (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 26893 | Lemma 2 for 2lgslem3 26896. (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 26894 | Lemma 3 for 2lgslem3 26896. (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 26895 | Lemma 4 for 2lgslem3 26896. (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 26896 | Lemma 3 for 2lgs 26899. (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 26897 | The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.) |
β’ (2 /L 2) = 0 | ||
Theorem | 2lgslem4 26898 | Lemma 4 for 2lgs 26899: special case of 2lgs 26899 for π = 2. (Contributed by AV, 20-Jun-2021.) |
β’ ((2 /L 2) = 1 β (2 mod 8) β {1, 7}) | ||
Theorem | 2lgs 26899 | 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 26806) 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 26900 | Lemma 1 for 2lgsoddprm 26908. (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))) |
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