Question: Is the following claim true?
Let $p$ be the odd prime then
$$\sum_{q=0}^{u}(n+qd)^{p-1}\ne p^t \ \ \ \ \forall n,u,d,t\in\mathbb{N}$$
also have claimed
$$\sum_{q=0}^{u}(n+qd)^{(p-1)m}\ne p^t \ \ \ \ \forall n,u,m,d,t\in\mathbb{N}$$
I apologize for the deleted update claim because I received some error. I will not let this happen again
Formula
$$\sum_{q=0}^{u}(n+qd)^{m}=\sum_{i=0}^{m} \binom{u+1}{i+1}\sum_{j=i}^{m}\binom{m}{j}n^{m-j}d^j\sum_{k=0}^{i}(i-k)^j(-1)^k\binom{i}k $$
Where $n,d\in \mathbb{R}$ and $u,m\in \mathbb{Z^*}$ and $0^0=1$
Proof : Formula for $\sum_{q=0}^{u}(n+qd)^{m}$
Related posts
Extending Fermat's Last Theorem
Can a sum of consecutive $n$th powers ever equal a power of two?
https://mathoverflow.net/q/348186/149083
I may not have tried much that you could reject using counter example