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A040036
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Primes p such that x^3 = 3 has a solution mod p.
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2
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2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 61, 67, 71, 73, 83, 89, 101, 103, 107, 113, 131, 137, 149, 151, 167, 173, 179, 191, 193, 197, 227, 233, 239, 251, 257, 263, 269, 271, 281, 293, 307, 311, 317, 347, 353, 359, 367, 383, 389, 401, 419, 431, 439, 443, 449, 461, 467, 479, 491, 499
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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MAPLE
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select(p -> isprime(p) and numtheory:-mroot(3, 3, p) <> FAIL, [2, seq(i, i=3..1000, 2)]); # Robert Israel, Nov 12 2017
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MATHEMATICA
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ok [p_]:=Reduce[Mod[x^3 - 3, p] == 0, x, Integers] =!= False; Select[Prime[Range[200]], ok] (* Vincenzo Librandi, Sep 11 2012 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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