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Number of triples (x,y,z) with x,y,z in the set {0,...,n-1} such that x*y*z is a cube and x^3 + 2*y^3 + 3*z^3 == 7 (mod n).
2

%I #14 Jun 24 2017 11:39:53

%S 1,4,7,10,14,18,1,22,27,29,34,37,35,7,47,46,54,56,44,60,17,71,71,72,

%T 76,60,80,15,92,93,81,98,104,109,25,115,118,90,91,121,129,33,114,142,

%U 135,142,145,147,22,153,163,121,168,176,174,29,135,182,185,184

%N Number of triples (x,y,z) with x,y,z in the set {0,...,n-1} such that x*y*z is a cube and x^3 + 2*y^3 + 3*z^3 == 7 (mod n).

%C Conjecture: For any positive integer n, the set {x^3 + 2*y^3 + 3*z^3: x,y,z = 0,...,n-1 and x*y*z is a cube} contains a complete system of residues modulo n.

%C See also A273488 for a similar conjecture.

%H Zhi-Wei Sun, <a href="/A273287/b273287.txt">Table of n, a(n) for n = 1..300</a>

%e a(2) = 4 since 0^3 + 2*0^3 + 3*1^3 == 7 (mod 2) with 0*0*1 = 0^3, 0^3 + 2*1^3 + 3*1^3 == 7 (mod 2) with 0*1*1 = 0^3, 1^3 + 2*0^3 + 3*0^3 == 7 (mod 2) with 1*0*0 = 0^3, and 1^3 + 2*1^3 + 3*0^3 == 7 (mod 2) with 1*1*0 = 0^3.

%e a(7) = 1 since 0^3 + 2*0^3 + 3*0^3 == 7 (mod 7) with 0*0*0 = 0^3.

%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]

%t Do[r=0;Do[If[Mod[x^3+2y^3+3z^3-7,n]==0&&CQ[x*y*z],r=r+1],{x,0,n-1},{y,0,n-1},{z,0,n-1}];Print[n," ",r];Continue,{n,1,60}]

%Y Cf. A000578, A273488.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Aug 28 2016