2016Forum MathematicumOpen access

On the smallest simultaneous power nonresidue modulo a prime

Kevin Ford, Moubariz Z. Garaev, Sergeĭ Konyagin

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Abstract

Abstract Let p be a prime and let p 1 , … , p r ${p_{1},\ldots,p_{r}}$ be distinct prime divisors of p - 1 ${p-1}$ . We prove that the smallest positive integer n which is a simultaneous p 1 , … , p r ${p_{1},\ldots,p_{r}}$ -power nonresidue modulo p satisfies n < p 1 / 4 - c r + o ⁢ ( 1 ) ( p → ∞ ) $n for some positive c r ${c_{r}}$ satisfying c r = e - ( 1 + o ⁢ ( 1 ) ) ⁢ r ${c_{r}=e^{-(1+o(1))r}}$ as r → ∞ ${r\to\infty}$ .

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Abstract Let p be a prime and let p 1 , … , p r ${p_{1},\ldots,p_{r}}$ be distinct prime divisors of p - 1 ${p-1}$ . We prove that the smallest positive integer n which is a simultaneous p 1 , … , p r ${p_{1},\ldots,p_{r}}$ -power nonresidue modulo p satisfies n < p 1 / 4 - c r + o ⁢ ( 1 ) ( p → ∞ ) $n for some positive c r ${c_{r}}$ satisfying c r = e - ( 1 + o ⁢ ( 1 ) ) ⁢ r ${c_{r}=e^{-(1+o(1))r}}$ as r → ∞ ${r\to\infty}$ .

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Available abstract

Abstract Let p be a prime and let p 1 , … , p r ${p_{1},\ldots,p_{r}}$ be distinct prime divisors of p - 1 ${p-1}$ . We prove that the smallest positive integer n which is a simultaneous p 1 , … , p r ${p_{1},\ldots,p_{r}}$ -power nonresidue modulo p satisfies n < p 1 / 4 - c r + o ⁢ ( 1 ) ( p → ∞ ) $n for some positive c r ${c_{r}}$ satisfying c r = e - ( 1 + o ⁢ ( 1 ) ) ⁢ r ${c_{r}=e^{-(1+o(1))r}}$ as r → ∞ ${r\to\infty}$ .

Key concepts: Modulo, Prime (order theory), Combinatorics, Prime power, Integer (computer science), Mathematics, Primitive root modulo n, Computer science

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