TITLE: Proving the congruence $p^{q-1}+q^{p-1} \equiv 1 \pmod{pq}$
QUESTION [10 upvotes]: If $p$ and $q$ are distinct primes , prove that $$p^{q-1}+q^{p-1} \equiv 1 \pmod {pq}$$


Using Fermat's Theorem we can get $$p^{q-1} \equiv 1 \pmod q, \qquad q^{p-1} \equiv 1 \pmod p.$$
After that I have no idea at all. Can anyone help me please

REPLY [3 votes]: To your
$$p^{q-1} \equiv 1 \pmod q\text, \qquad q^{p-1} \equiv 1 \pmod p\text,$$
add the fact that
$$p^{q-1} \equiv 0 \pmod p\text, \qquad q^{p-1} \equiv 0 \pmod q\text.$$
Then
$$p^{q-1} + q^{p-1} \equiv 1 \pmod q\text, \qquad p^{q-1} + q^{p-1} \equiv 1 \pmod p\text,$$
and by the Chinese Remainder Theorem,
$$p^{q-1} + q^{p-1} \equiv 1 \pmod {pq}\text.$$