How to prove $\gcd(a^m-b^m,a^n-b^n) = a^{\gcd(m,n)} - b^{\gcd(m,n)} $? A problem taken from the exercises of Concrete Mathematics by Graham, Knuth, and Patashnik  is as follows: 

Prove that if $a\perp b$ and $a>b$ then
     $$\gcd(a^m-b^m,a^n-b^n) = a^{\gcd(m,n)} - b^{\gcd(m,n)},\quad 0\leq m<n.$$

(In the notation of the book, $a\perp b$ means that $a$ and $b$ are relatively prime, i.e. $\gcd(a,b)=1$.)
I can not prove this equation. Can you please help me to prove this formula?  
 A: This is exercise 4.38. There is a hint to use Euclid's algorithm that you forgot to reproduce. There is also an answer (p. 503) that reads
$a^n-b^n =(a^m-b^m)(a^{n-m}b^0+a^{n-2m}b^m+\cdots+a^{n\bmod m}b^{n-m-n\bmod m})+b^{m\lfloor n/m\rfloor}(a^{n\bmod m}-b^{n\bmod m})$
What this means is that the first step of Euclid's algorithm reduces $\gcd(a^n-b^n,a^m-b^m)$ to $\gcd(a^m-b^m,b^{m\lfloor n/m\rfloor}(a^{n\bmod m}-b^{n\bmod m}))$. But $b^{m\lfloor n/m\rfloor}$ is relatively prime to $a^n-b^n$ since it divides the second term and is relatively prime to the first term; therefore it will be relatively prime to the $\gcd$ that is being computed, and we might as well remove that factor from the second argument of the $\gcd$. All in all this gives
$$
  \gcd(a^n-b^n,a^m-b^m)
  =\gcd(a^m-b^m,a^{n\bmod m}-b^{n\bmod m}).
$$
Now iterating as in the Euclidean algorithm eventually gives
$$
\gcd(a^m-b^m,a^n-b^n) = a^{\gcd(m,n)} - b^{\gcd(m,n)}.
$$
A: Outline of a proof:
Let $d\mid a^n-b^n$ and $d\mid a^m-b^m$.  
The "slick" solution is to show that if $mx+ny=\gcd(m,n)$, then since:
$$a^m\equiv b^m\pmod d$$
and 
$$a^n\equiv b^n\pmod d$$
Then:
$$a^{(m,n)}=(a^m)^x (a^n)^y \equiv (b^m)^x(b^n)^y = b^{(m,n)}\pmod d$$
The only tricky part here is that you need to understand that we can invert $a$ and $b$ modulo $d$ since $a\perp b$ means that $d\perp a$ and $d\perp b$.  So, although $x$ or $y$ might be negative, you can work around this by taking care.
