User:Vinkmar

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m363a4q7

$p^{q-1}+q^{p-1}\equiv 1(\mathrm{mod}pq)$

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

$p^{q-1}+q^{p-1}\equiv 1(\mathrm{mod}p)$ and $p^{q-1}+q^{p-1}\equiv 1(\mathrm{mod}q)$

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

$p^{q-1}\equiv 0(\mathrm{mod}p)$ and $q^{p-1}\equiv 1(\mathrm{mod}p)$.

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