Show that $a^{16}-b^{16}$ is divisible by $133$ if $a$ and $b$ are both prime to $85$
Since $(85, a)=1(17,5)$ and $(85, b)=(17,5)$ then $a^{16}-1\equiv (mod ~17)$, $a^{4}-1\equiv (mod~ 5)\implies a^{16}-1\equiv (mod~ 5)$, $b^{16}-1\equiv (mod~ 17)$, $b^{4}-1\equiv (mod~ 17)\equiv b^{16}-1\equiv (mod~ 5)$. Then $a^{16}-1\equiv (mod~ 85)$ $a^{16}-1\equiv (mod~ 85)$ Thus $a^{16}-b^{16}\equiv 0 (mod~85)$.
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