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?

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

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 $p^n \not\equiv p \bmod p^2$.


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