Syndrome: $ a> b $ and $ gcd (a, b) = 1 $, there $ gcd (a ^ nb ^ n, a ^ mb ^ m) = a ^ {gcd (n, m)} - b ^ { gcd (n, m)} $.
prove:
Suppose $ n> m $, $ r = n \% m $.
The Euclidean algorithm,
$a^n - b^n = (a^m-b^m)(a^{n-m} + a^{n-2m}b^m + ...+) + a^rb^{n-r} - b^n$，
$gcd(a^n-b^n, a^m-b^m) = gcd(a^m-b^m, a^rb^{n-r}-b^n) = gcd(a^m-b^m, b^{n-r}(a^r-b^r))$，
Because $ r = n \% m $, so $ b ^ {nr} = b ^ {m \ left \ lfloor \ frac {n} {m} \ right \ rfloor} = b ^ {km} $.
Consider $ gcd (b ^ {km}, a ^ mb ^ m) $,
A polynomial division by a $ b ^ {km} = (a ^ mb ^ m) (- b ^ {(k-1) m} - a ^ mb ^ {(k-2) m} -...- a ^ { (k-1) m}) + a ^ {km} $,
$gcd(b^{km}, a^m-b^m) = gcd(a^{km}, a^m-b^m) = d$，
$ D | b ^ {km}, \ d | a ^ {km}, \ d | gcd (b ^ {km}, a ^ {km}) = 1 $, so $ d = 1 $, i.e. $ GCD ( b ^ {nr}, a ^ mb ^ m) = 1 $.
所以 $gcd(a^n-b^n, a^m-b^m) = gcd(a^m-b^m, a^{n \% m}-b^{n \% m}) = a^{gcd(n,m)} - b ^ {gcd(n,m)}$.
(In fact, the whole process is the Euclidean algorithm)