Hi, I have this question,

Show that for any integer n, $\displaystyle n^{13}\equiv{n}(\bmod7)$

I have written this:

$\displaystyle n^{6}\equiv{1}(\bmod7)$

because of Eulers theorem.

$\displaystyle n^{13} = (n^6)^2n$

So $\displaystyle (n^6)^2\equiv{1}(\bmod7)$

So $\displaystyle n\equiv{n}(\bmod7)$

And $\displaystyle 0\equiv{0}(\bmod7)$ QED

Is this right? Please help.

Thanks,

Katy