OFFSET
1,1
COMMENTS
Being a cube mod p is a necessary condition for 3 to be a 9th power mod p. See Williams link pp. 1, 8 (warning: term 271 is missed). - Michel Marcus, Nov 12 2017
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Kenneth S. Williams, 3 as a Ninth Power (mod p), Math. Scand. Vol 35 (1974), 309-317.
MAPLE
select(p -> isprime(p) and numtheory:-mroot(3, 3, p) <> FAIL, [2, seq(i, i=3..1000, 2)]); # Robert Israel, Nov 12 2017
MATHEMATICA
ok [p_]:=Reduce[Mod[x^3 - 3, p] == 0, x, Integers] =!= False; Select[Prime[Range[200]], ok] (* Vincenzo Librandi, Sep 11 2012 *)
PROG
(Magma) [p: p in PrimesUpTo(450) | exists(t){x : x in ResidueClassRing(p) | x^3 eq 3}]; // Vincenzo Librandi, Sep 11 2012
(PARI) isok(p) = isprime(p) && ispower(Mod(3, p), 3); \\ Michel Marcus, Nov 12 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved