Question

Show that $n^{36}-1$ is divisible by $33744$ if $n$ is prime to $2,3,19$ and $37$.

Answer 1

$n^{36}-1$

if $p$ is a prime number and $N$ is prime to $p$ then $N^{p-1}-1$ is a multiple of $p$

Pu $n=37$

$\Rightarrow n^{37-1}-1=n^{36}-1$ is divisible by $\mathrm{37......}\left(i\right)$

$n^{36}-1=(n^{18}-1)(n^{18}+1)$

Again using fermats theorem

Put $n=19$

$\Rightarrow n^{19-1}-1=n^{18}-1$ is divisible by $\mathrm{19.....}\left(ii\right)$