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**ilikecandy** show that if an integer n has at least two distinct odd prime divisors then there exists $\displaystyle k < \varphi(n)$ such that $\displaystyle a^k \equiv 1 \mod n,$ for every a relatively prime to n.

What I have so far: The expression a^k ≡ 1(mod n) looks similar to Euler's Generalization of Fermat's Little Theorem. Also, n can be expressed as xp1p2, where x is a positive integer, and p1 and p2 are odd distinct prime numbers.

I know this isn't much, but I don't to where to go afterwards to complete the proof.