2
$\begingroup$

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

$\endgroup$
8
  • 1
    $\begingroup$ 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, 2019 at 10:54
  • 1
    $\begingroup$ Finished with no counter example, now I check upto $100$ $\endgroup$
    – Peter
    Dec 12, 2019 at 11:06
  • $\begingroup$ @Peter thank you so much, can you share your algorithm $\endgroup$
    – Pruthviraj
    Dec 12, 2019 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, 2019 at 12:47
  • $\begingroup$ Still no counterexample, but range $100$ not yet finished $\endgroup$
    – Peter
    Dec 12, 2019 at 12:48

1 Answer 1

0
$\begingroup$

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 $p\neq 2$.
$\endgroup$
5
  • $\begingroup$ Thanks, what can we conclude from this. $\endgroup$
    – Pruthviraj
    Dec 15, 2019 at 5:13
  • 1
    $\begingroup$ 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, 2019 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, 2019 at 9:26
  • 1
    $\begingroup$ I don't know, however I'd teach you what I know of PARI GP. $\endgroup$
    – user645636
    Dec 16, 2019 at 15:00
  • 1
    $\begingroup$ @Pruthviraj if you want help learning just invite me to a chat. $\endgroup$
    – user645636
    Dec 21, 2019 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.