I am seeking a proof for the following...
Suppose $p$ and $q$ are distinct primes. Show that $$ p^{q-1} + q^{p-1} \equiv 1 \quad (\text{mod } pq)$$ $$$$ $$$$ I gather from Fermat's Little Theorem the following: $$q^{p-1} \equiv 1 \quad (\text{mod } p)$$
and
$$p^{q-1} \equiv 1 \quad (\text{mod } q)$$
How can I use this knowledge to give a proof? I'm confident I can combine this with the Chinese Remainder Theorem, but I am stuck from here.
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