Question

Prove that $n^{16}-1$ is divisible by 17, if $(n,17)=1.$

Answer 1

$(n,17)=1$ and 17 is a prime number.
$\therefore$By Fermat theorem

$n^{17-1}\equiv 1mod17$

If $p$ is prime and $(a,p)=1$, then $a^{p-1}\equiv 1modp$

or $n^{16}\equiv 1mod17$ or $17|(n^{16}-1)$

i.e., $(n^{16}-1)$ is divisible by 17.