Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$ 
Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.

This is what I have done so far:
Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.
I am trying to create a linear combination of $a^2$ and $b^2$ but do not know what to do with the middle term.
EDIT: I would be interested in seeing a proof using the Fundamental Theorem of Arithmetic as well. I simply thought it would be easiest to use Bezout's Identity. 

Hint: Greatest common divisor will always have the minimum of the exponents(from the $2$ number) from the prime factorization. The minimum exponents of $a^2$ and $b^2$ are the same like the minimal exponents of $a$ and $b$ multiplied by $2$.
