A374592 Numbers k such that 3*k^4 - 3*k^2 + 1 is prime.

2, 5, 7, 8, 9, 14, 15, 20, 23, 30, 36, 37, 43, 48, 49, 50, 54, 56, 57, 69, 71, 79, 85, 86, 91, 93, 97, 98, 106, 111, 112, 119, 124, 128, 131, 133, 134, 135, 140, 154, 159, 162, 167, 180, 181, 198, 204, 208, 212, 226, 232, 236, 246, 259, 278, 281, 285, 286, 288

Offset: 1

Author

Jon E. Schoenfield, Jul 12 2024

Keywords

• nonn

Comments

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.