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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

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    $\begingroup$ 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. $\endgroup$ – Peter Dec 12 '19 at 10:54
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    $\begingroup$ Finished with no counter example, now I check upto $100$ $\endgroup$ – Peter Dec 12 '19 at 11:06
  • $\begingroup$ @Peter thank you so much, can you share your algorithm $\endgroup$ – Pruthviraj Dec 12 '19 at 11:39
  • $\begingroup$ 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]))))))) $\endgroup$ – Peter Dec 12 '19 at 12:47
  • $\begingroup$ Still no counterexample, but range $100$ not yet finished $\endgroup$ – Peter Dec 12 '19 at 12:48
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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 $p\neq 2$.
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  • $\begingroup$ Thanks, what can we conclude from this. $\endgroup$ – Pruthviraj Dec 15 '19 at 5:13
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    $\begingroup$ well if my above code gave consistent values on reordering tests I'd say just 60719 pairs (n,d) in the first million pass these criteria. $\endgroup$ – user645636 Dec 15 '19 at 11:47
  • $\begingroup$ Actually I need some time to learn pari\gp and focus on given post. Can you please comment, are my claims false or true? $\endgroup$ – Pruthviraj Dec 16 '19 at 9:26
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    $\begingroup$ I don't know, however I'd teach you what I know of PARI GP. $\endgroup$ – user645636 Dec 16 '19 at 15:00
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    $\begingroup$ @Pruthviraj if you want help learning just invite me to a chat. $\endgroup$ – user645636 Dec 21 '19 at 21:02

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