I am seeking a proof for the following…
Suppose $p$ and $q$ are distinct primes. Show that $$
p^{q-1} + q^{p-1} \equiv 1 \quad (\text{mod } pq)$$
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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.
Best Answer
Write $a=p^{q-1}+q^{p-1}$. Immediately we have
$$\begin{cases} a\equiv 0^{\,q-1}+q^{p-1} \equiv 1 \mod p \\ a\equiv p^{q-1}+0^{\,p-1} \equiv 1 \mod q.\end{cases}$$
What does CRT tell us $a$ must be $\bmod pq$?