Vinkmar

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m363a4q7

$p^{q-1}+q^{p-1}\equiv 1{\pmod {pq}}$

But $p$ and $q$ are distinct primes, so for the above to be valid, the following two equations must hold:

$p^{q-1}+q^{p-1}\equiv 1{\pmod {p}}$ and $p^{q-1}+q^{p-1}\equiv 1{\pmod {q}}$

Considering only the first of the two equations (the latter case is, for lack of a better term, symmetrical), we have:

$p^{q-1}\equiv 0{\pmod {p}}$ and $q^{p-1}\equiv 1{\pmod {p}}$.

The former is obviously true. The latter is proven using Fermat's Little Theorem, QED