I am trying to prove a little modification on Euler's theorem

$\displaystyle a^{\phi(n)+1}\equiv a (mod n)$, for all a, and where n is the product of two distinct primes say p and q.

I am having problems with the for all a part as the above is easy to show.

$\displaystyle a^{\phi(n)} \equiv 1 (mod n)$

$\displaystyle a * a^{\phi(n)} \equiv a (mod n)$

$\displaystyle a^{\phi(n) + 1} \equiv a (mod n)$

I need to show that if q or p is a factor of a, then it still yields the same result, as Euler's theorem says that the gcd(a, n) must be 1. But I am not sure how to show that in the modification it doesn't matter. Any ideas?

Thanks!