# 57156 binomials

...
• 17
• 163
• 0

## Báo cáo toán học: "The Number of Permutation Binomials Over F4p+1 where p and 4p + 1 are Primes" doc

... all permutation binomials occur in pairs over F 4p+ 1 where p and 4p + are primes Theorem 12 Let q = 4p + where p and q are primes The following are paired permutation binomials over Fq : (i) (1, ... < 2p then (mod 4) m 1 m≡2 m≡3 m≡4 p 1 p 1 N2 N2 N1 N2 N1 N1 N2 N1 (2) If 2p < m < 3p then (mod 4) m 1 m≡2 m≡3 m≡4 p 1 p 1 2 (p 1) + N2 2 (p 1) + N2 N1 N2 2 (p 1) + N1 2 (p 1) + N1 N2 ... t=0 2p if p 1 2t +1 (a 1 )2 (p t) 1 = Indeed, this follows from t=0 (1 + a) 2p (1 − a) 2p = ⇐⇒ (a 1 + 1) 2p − (a 1 1) 2p = ⇐⇒ (1 + a 1 ) 2p (1 − a 1 ) 2p = Theorem 12 reduces the problem of counting...
• 15
• 157
• 0

## Báo cáo toán học: " Nonexistence of permutation binomials of certain shapes" ppsx

... modulo q − Permutation binomials in general Lemma 2.1 If f is a permutation polynomial over Fq , then the greatest common divisor of the degrees of the terms of f is coprime to q − Proof Otherwise ... (mod 4) The set of degrees of terms of f +1 is S = {n + n, n + n + 4, n + n + 8, , n + n + + 4} Since S consists of + consecutive multiples of 4, it certainly contains a multiple of , so (by Hermite’s ... Now the set of degrees of terms of f +2 is S = {2 n + 2n, n + 2n + 4, n + 2n + 8, , n + 2n + 4(2 + 2)} Here S consists of + consecutive multiples of 4, so it contains a multiple of p − = By...
• 5
• 73
• 0

Xem thêm