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Numbers k such that 3*k^4 - 3*k^2 + 1 is prime.
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%I #6 Jul 15 2024 03:33:02

%S 2,5,7,8,9,14,15,20,23,30,36,37,43,48,49,50,54,56,57,69,71,79,85,86,

%T 91,93,97,98,106,111,112,119,124,128,131,133,134,135,140,154,159,162,

%U 167,180,181,198,204,208,212,226,232,236,246,259,278,281,285,286,288

%N Numbers k such that 3*k^4 - 3*k^2 + 1 is prime.

%C Equivalently, numbers k such that there exists a prime of the form k^6 - m^3. Proof: Let d = k^2 - m. Then m = k^2 - d, so k^6 - m^3 = k^6 - (k^2 - d)^3 = k^6 - (k^6 - 3*k^4*d + 3*k^2*d^2 - d^3) = d*(3*k^4 - 3*k^2*d + d^2), which cannot be prime unless d = 1, i.e., k^6 - m^3 = 3*k^4 - 3*k^2 + 1.

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Jul 12 2024