I noticed that $\gcd(a^x,b^y)=\gcd(a,b)^z$ for a, b, x, y and z positive integers,
My question is how to prove that?
Additionally assuming the above is correct, when a and b are relatively prime (co-prime):
As $\gcd(a,b)=1$, $\gcd(a^x,b^y)=\gcd(a,b)^z=1^z=1$
Does this means that $a^x$ and $b^y$ are co-primes too in this case?
Update:
I come to found a counterexample
$\gcd(8^3,4^5)=512$, $\gcd(8,4)=4$ and there is no z positive integer that $4^z=512$
The claim is always true if $\gcd(a,b)=1$ or if $\gcd(a,b)=p$ where $p$ is prime. This is because if $\gcd(a^x,b^y)$ is not a power of $p$ then there is some prime $q\neq p$ which divides $\gcd(a^x,b^y)$ and hence divides $a^x$ and $b^y$. But since $q$ is prime, $q\mid a^x\Longrightarrow q\mid a$, etc., so $q\mid\gcd(a,b)$, which is a contradiction.
For any number $h$ which is neither $1$ nor a prime there are values $a,b$ for which $\gcd(a,b)=h$ but $h<\gcd(a^2,b)<h^2$, giving a counterexample: take $a=h, b=ph$ for some prime $p\mid h$.