I need to prove for a given n, if $\phi(x)=n$ has a solution for x, it always has another?
We know $\phi(2)=\phi(1)=1$ and can easily prove that n must be even for x>2.
So, n can be of the form $2^a.q$ where a>0 , odd q are natural numbers.
If you need to prove that, you're in big trouble. It's Carmichael's conjecture, and it's wide open.