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%I #10 Jan 29 2019 04:38:02
%S 151,241,251,431,571,641,911,971,1181,1811,2011,2351,2381,2411,2731,
%T 3061,3121,3221,3251,3301,3331,3361,3391,3541,3761,3821,3881,4211,
%U 4751,4861,4871,4931,5021,5381,5441,5471,5581,5641,5711,5821,5861
%N Primes p such that x^5 == 2 (mod p) has five solutions.
%C For any prime modulus, there must be exactly 0, 1 or 5 solutions to the equation with x between 0 and p - 1.
%C Primes == 1 (mod 5) such that 2 is a quintic residue, that is, primes p such that 2^((p-1)/5) == 1 (mod p). - _Jianing Song_, Jan 27 2019
%o (PARI) forstep(p=11, 5000, 10, if(isprime(p)&&Mod(2,p)^((p-1)/5)==1, print1(p, ", "))) \\ _Jianing Song_, Jan 27 2019
%Y Cf. A040159.
%K nonn
%O 1,1
%A _Don Reble_, Sep 20 2001
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