1
$\begingroup$

I am working on a problem in number theory:

Problem: Let $p>2$ be prime. Let $n$ be an integer with $\gcd(n,p)=1$ and $n\not\equiv1\pmod{p}$. Prove that $p\mid(1+n+n^2+n^3+\dots+n^{p-2})$.

Right now, I am trying to solve this problem with Fermat's Little Theorem. The Theorem tells me $n^p\equiv n\pmod{p}$. Then I have $n(n^{p-1}-1)\equiv0\pmod p$, so I know $p\mid n(n^{p-1}-1)$. Because $p\not\mid n$, I know then that $p\mid(n^{p-1}-1)$. From here, I am stuck and cannot see how to proceed. Thank you in advance for assistance.

$\endgroup$
1
  • 3
    $\begingroup$ Hint: first consider whether $p\mid(1+n+n^2+n^3+\dots+n^{p-2})(1-n)$. $\endgroup$ – Greg Martin Feb 2 at 17:13
1
$\begingroup$

Since $n\not\equiv1\pmod{p} \implies n \neq pk+1 \implies n-1 \neq pk \implies p \nmid n-1$.

As Greg Martin mentions, consider $(1+n+n^2+n^3+\dots+n^{p-2})(1-n)=1-n^{p-1}$.

By Fermat's Little Theorem, for any prime $p$ we have $n^{p-1} \equiv 1 \pmod p \implies p\mid 1-n^{p-1} \implies p \mid (1+n+n^2+n^3+\dots+n^{p-2})(1-n).$

But $p \nmid n-1$, so $p\nmid 1-n$ and $p \mid(1+n+n^2+n^3+\dots+n^{p-2})$.

$\endgroup$

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.