User: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