
COMMENTS

a(4) = 0. Proof: For k == 1 (mod 4), 4^k + k^4 = 4*x^4 + k^4 = (2*x^2  2*k*x + k^2)(2*x^2 + 2*k*x + k^2), where x = 4^((k1)/4). For k == 3 (mod 4), 4^k + k^4 = 64*x^4 + k^4 = (8*x^2  4*k*x + k^2)(8*x^2 + 4*k*x + k^2), where x = 4^((k3)/4) (cf. Israel, 2015).
Conjecture: a(6) = 0.
From Jon E. Schoenfield, Nov 13 2018: (Start)
Let t = 6^k + k^6.
If k is even, then 2t.
If k is odd but not divisible by 7, then 7t.
If k is divisible by 3, then 3t.
If k == 7 or 63 (mod 70), then 5t.
Thus, a(6) == 35, 49, 91, 119, 161, or 175 (mod 210) if a(6) > 0. (End)
