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. 
 A: Suppose that $(a,b)=d$. Then Bezout's Identity says that we have some $x,y$ so that $ax+by=d$, and therefore
$$
a^3x^3+3a^2bx^2y+3ab^2xy^2+b^3y^3=d^3\tag{1}
$$
Dividing $(1)$ by $d$, remembering that both $d\mid a$ and $d\mid b$, we get
$$
a^2\left(\frac adx^3+3\frac bdx^2y\right)+b^2\left(3\frac adxy^2+\frac bdy^3\right)=d^2\tag{2}
$$
$(2)$ says that $\left.\left(a^2,b^2\right)\middle|\,d^2\right.$. Since $d\mid a$ and $d\mid b$, we also have $\left.d^2\,\middle|\left(a^2,b^2\right)\right.$.
Therefore,
$$
\left(a^2,b^2\right)=d^2=(a,b)^2
$$
A: 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 $2$.
A: $(a,b)$ is the usual notation for $\gcd(a,b)$.
Remember $\, n\mid a,b\iff n\mid (a,b)\,$ by definition of $\gcd$.
Use distributive property $\,(ak,bk)=k(a,b)$.    
As Bill says, you can prove it in this similar way:   
$c\mid (ak,bk)\iff c\mid ak,bk\iff \frac{c}{k}\mid a,b\iff\frac{c}{k}\mid (a,b)\iff c\mid k(a,b)$
Another lemma: $((a,b),c)=(a,b,c)$.   
Proof: $(a,b,c)\mid a,b,c\iff (a,b,c)\mid (a,b),c\iff (a,b,c)\mid ((a,b),c)$   
$((a,b),c)\mid (a,b),c\iff ((a,b),c)\mid a,b,c\iff ((a,b),c)\mid (a,b,c)\ \ \ \square$    
Prove $(a,b)(a^2,b^2)=(a,b)^3$, then divide by $(a,b)$ to finish your proof.  
$(a,b)(a^2,b^2)=(a(a^2,b^2),b(a^2,b^2))=((a^3,ab^2),(a^2b,b^3))=(a^3,a^2b,ab^2,b^3)$
$(a,b)(a,b)=((a,b)a,(a,b)b)=((a^2,ab),(ab,b^2))=(a^2,ab,b^2)$  
$(a^2,ab,b^2)(a,b)=(a^2(a,b),ab(a,b),b^2(a,b))=((a^3,a^2b),(a^2b,ab^2),(ab^2,b^3)),$   
which too is $(a^3,a^2b,ab^2,b^3)$. $\ \ \ \square$   
