# Not represent prime power?

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 $$0^0=1$$

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

• I am currently checking $n,u,d,t$ in the range $[1,50]$ and $p$ in the range $[3,50]$ without finding a counter example yet. – Peter Dec 12 '19 at 10:54
• Finished with no counter example, now I check upto $100$ – Peter Dec 12 '19 at 11:06
• @Peter thank you so much, can you share your algorithm – Pruthviraj Dec 12 '19 at 11:39
• PARI/GP : for(n=1,100,for(u=1,100,for(d=1,100,for(t=1,100,forprime(p=3,100,if(sum(q=0,u,(n+q*d)^(p-1))==p^t,print([n,u,d,t,p]))))))) – Peter Dec 12 '19 at 12:47
• Still no counterexample, but range $100$ not yet finished – Peter Dec 12 '19 at 12:48

• $$\gcd(n,d)$$ 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 $$p\neq 2$$.