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In short words: when you compute things modulo $n = pq$, you are really computing things simultaneously modulo $p$ and modulo $q$. That's the gist of the Chinese Remainder Theorem. So to prove that $a = b \pmod n$, you just have to prove that $a = b \pmod p$ and $a = b \pmod q$. Modulo $p$, for any $x$ that is not a multiple of $p$, $x^{p-1} = 1 \pmod p$ ...

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