login
Primes p such that x^3 = 15 has a solution mod p.
3

%I #12 Sep 08 2022 08:44:53

%S 2,3,5,7,11,17,23,29,31,41,47,53,59,67,71,79,83,89,101,107,113,131,

%T 137,149,167,173,179,191,197,223,227,229,233,239,251,257,263,269,277,

%U 281,283,293,311,317,331,347,353

%N Primes p such that x^3 = 15 has a solution mod p.

%C Complement of A040070 relative to A000040. - _Vincenzo Librandi_, Sep 13 2012

%H Vincenzo Librandi, <a href="/A040069/b040069.txt">Table of n, a(n) for n = 1..1000</a>

%t ok [p_]:=Reduce[Mod[x^3 - 15, p] == 0, x, Integers] =!= False; Select[Prime[Range[180]], ok] (* _Vincenzo Librandi_, Sep 11 2012 *)

%o (Magma) [p: p in PrimesUpTo(450) | exists(t){x : x in ResidueClassRing(p) | x^3 eq 15}]; // _Vincenzo Librandi_, Sep 11 2012

%o (PARI) select(n->ispower(Mod(15, n),3), primes(500)) \\ _Charles R Greathouse IV_, Sep 11 2012

%Y Cf. A000040, A040070.

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_.