Proving $133|a^{18}-b^{18}$ if $\gcd(a,133)=\gcd(b,133)=1$.

Question

If $\gcd(a,133)=\gcd(b,133)=1$ then prove that $133|a^{18}-b^{18}$.

Using Fermat theorem: $a^{132} \equiv 1\mod\ 133$ and $b^{132} \equiv 1\mod\ 133$, so $a^{132} \equiv b^{132}\mod\ 133$. What should I do further.

Answer 1

$133=7\cdot19$ is not prime, so you can't use Fermat's little theorem this way.

You have $a^6 \equiv 1 \mod 7$ and $a^{18} \equiv 1 \mod 19$ by Fermat's little theorem instead, since $\gcd(a,133)=1$.

So $a^{18} \equiv 1 \mod 7$ and $a^{18} \equiv 1 \mod 19$. Using the Chinese remainder theorem, $a^{18} \equiv 1 \mod 133$. Similiarly, $b^{18} \equiv 1 \mod 133$, since also $\gcd(b,133)=1$, so $$a^{18}-b^{18} \equiv 1-1 =0 \mod 133$$

Hence $133 \mid a^{18}-b^{18} $ for all $a,b$ such that $\gcd(a,133)=1$ and $\gcd(b,133)=1$.