GCD of two polynomials in Mod 2 
Possible Duplicate:
Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ 

Let $p$ and $q$ be distinct primes.
I wonder is the following statement always true?
$$\gcd(x^p-1, x^q-1) \stackrel{?}{=} x-1$$

More generally, we have $\gcd(x^n-1,x^m-1)=x^{\gcd(n,m)}-1$.
