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Bernoulli polynomials and determine the necessary and sufficient conditions for prime numbers


Lemma 4 and ( 6) have an odd number of P is a prime time when the p-lJ'2m , (2m +1) n a 11I2rn time , (2m +1) S2m (-) S2P - 12 ( Dan j. ( rood (2m +1) p); when p) (mod (2m ten 1) P) when the bone when the P- 152m B + () 0 (m. d (2m +1) p); when a 1I2m when p , B (p_) () (mod (2m ten 1) p). In addition , the use of Lemma 2,[link widoczny dla zalogowanych], Lemma 3 and Theorem 1 , easy to get completely similar to Theorem 4 P is a prime odd number of necessary and sufficient condition is met ( 1) 2 ≤ 2m ≤ guide have a 1 B (p) =- -0 (Rood (2m a 1) p) (2) B (p - 1); a 2p (modp) Theorem j P is a prime odd number of necessary and sufficient condition is met ( 1) 2 ≤ 2m ≤ a guide 1 has B. + () 0 (rood (2nI1) p) (2) B () A p (p a 1) (modp) 3 living plus the equivalent propositions we guess [j], [6] to get the session plus conjecture Guess guess one of the three equivalent P is a prime odd number of necessary and sufficient conditions , SH ( Dan ); (modp) P is a prime suspect 2 odd number of necessary and sufficient condition , S (p - 1) b 1) (modp ) 18 Guangxi Teachers University (Natural Science Edition) No. l6 guess Volume 3 P is a prime odd number of necessary and sufficient conditions , PI (pB a +1) elements here , the use of Theorem 1 and above conjecture , obviously increases the availability of home conjecture The other three equivalent conjecture : Conjecture 4, P is a prime odd number of necessary and sufficient condition , Bp (p -1 ) a 2p (roodP) P is a prime suspect 5 odd number of necessary and sufficient condition is. B () a p_ (modp)'' conjecture P is a prime odd number 6 necessary and sufficient condition , 2Bp + l (p) c 2Bp + l-P. (roodP) we only give proof of the equivalence of Conjecture 6 : from (1 ) may Bp +1 (p) a B + a (p +1) S (p a 1) (p is odd) (13) by the conjecture 2 was the odd prime P Velvet 2s (p-1) a P (modp3) Velvet 2 (p +1) S (p-1)-p (p +1) Lan a P (roodpS) Velvet 2BD + 】 (p ) Blue 2B mouth + l - p (modp3) Thus, to prove or disprove the Home plus conjecture , to prove or disprove any of the above conjecture to conjecture , but is still very difficult , yet continue to study . [13 (4) [2] [3] E43E5] Es] [7]