For k ≡ 2 mod 4 the equation x^k + (x+1)^k = y^n (n ≥ 3) has only the solutions x = 0, -1 when 6 ≤ k ≤ 100 or k has an odd prime factor ≡ 3 mod 4.
Bennett and Chris M
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Sum of consecutive powers as a perfect power
For k ≡ 2 mod 4 the equation x^k + (x+1)^k = y^n (n ≥ 3) has only the solutions x = 0, -1 when 6 ≤ k ≤ 100 or k has an odd prime factor ≡ 3 mod 4.