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A336450
Values z of primitive solutions (x, y, z) to the Diophantine equation x^3 + y^3 + 2*z^3 = 2*5^6.
1
1, -3, 4, 9, 16, 25, 36, 49, -56, 64, 81, 88, -104, 121, 144, -167, 169, 177, 196, -203, -243, -255, 256, 277, 289, 324, 361, -363, 373, -395, -411, 441, 484, 529, 576, 676, 709, -719, 729, 784, 841, 961, 1017, 1024, -1028, 1080, 1089, -1091, 1156, 1296, 1369
OFFSET
1,2
COMMENTS
Terms are arranged in order of increasing absolute value (if equal, the negative number comes first).
Let x = a^(2*m) - (a^m)*t - t^2, y = a^(2*m) + (a^m)*t - t^2, z = t^2; then x^3 + y^3 + 2*z^3 = 2*a^(6*m). When a = 5, m = 1, t = 5*n + k(k = {1, 2, 3, 4}); (x, y, z) are primitive solutions of equation. Thus, A047201(n)^2 are terms of the sequence.
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, D5.
EXAMPLE
(-20)^3 + 34^3 + 2*(-3)^3 = 31250, -3 is a term.
(-11)^3 + 29^3 + 2*16^3 = 15^3 + 27^3 + 2*16^3 = 31250, 16 is a term.
MATHEMATICA
Clear[t]
t = {};
Do[y = (31250 - x^3 - 2z^3)^(1/3) /. (-1)^(1/3) -> -1;
If[IntegerQ[y] && GCD[x, y, z] == 1, AppendTo[t, z]], {z, -1369, 1369}, {x, -Round[(Abs[31250 - 2z^3]/3)^(1/2)], Round[(Abs[31250 - 2z^3]/3)^(1/2)]}]
u = Union@t;
v = Table[(-1)^n*Floor[(n + 1)/2], {n, 0, 2739}];
Select[v, MemberQ[u, #] &]
KEYWORD
sign
AUTHOR
XU Pingya, Aug 08 2020
STATUS
approved