0
$\begingroup$

Show that there does not exist any $n$ for which $a^n=a$ forall $a\in \Bbb Z_m$ where $m$ is divisible by the square of some prime.

Suppose such $n$ exists and $m=p^2k$ for some prime $k$.

then $a^n=a$. Also $ma=0\implies p^2k a=0$

How to derive a contradiction from here?

$\endgroup$
  • 2
    $\begingroup$ what about $n=1$? $\endgroup$ – Vasya Oct 5 '17 at 20:03
1
$\begingroup$

The statement

There does not exist $n>1$ such that $a^n \equiv a \bmod m$ for all $a \in \mathbb Z$

is equivalent to

For all $n>1$, there is $a \in \mathbb Z$ such that $a^n \not\equiv a \bmod m$

It is enough to prove this for $m=p^2$, with $p$ prime.

Indeed, if $n>1$, then $a=p$ works because $p^n \equiv 0 \bmod p^2$ and so $p^n \not\equiv p \bmod p^2$.

$\endgroup$

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.