Solutions of $\phi(x)=n$ for a given n.

Question

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.

Answer 1

If you need to prove that, you're in big trouble. It's Carmichael's conjecture, and it's wide open.

Answer 2

If $n$ is odd then $\phi(n)=\phi(2n)$.