Not represent prime power?

Question

Question: Is the following claim true?

Let $p$ be the odd prime then

$$\sum_{q=0}^{u}(n+qd)^{p-1}\ne p^t \ \ \ \ \forall n,u,d,t\in\mathbb{N}$$

Proof for $p=3$

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

Answer 1

Here's what I know:

• $\gcd(n,d)$ has to divide every term of the sum, and therefore the sum itself.
• For the sum to be a prime power, that means $\gcd(n,d)$ is 1,p, or $p^x~~~~x\leq t~~~~x,t\in\mathbb{N}$
• By Fermat's little theorem, we get that all terms except the ones that have a factor of $p$ will have remainder 1, on division by $p$ ( there needs to be a multiple of $p$ of these in the sum for it to work out).
• Because the sum of two odd multiples, is an even multiple of any given number; we know that an odd number of terms must exist, that are odd multiples of the gcd above; any time $p\neq 2$.