prove that if $\gcd(a,b)=1$ then $\gcd(a^2, b^2)=1$

Using Bezout's identity I have shown

$$(a,b^2)=(b,a^2)=1$$

but what should be the next step? Thanks.

• HINT: for any prime $p$ , $p|\gcd(a^2,b^2)$, then $p^2|\gcd(a^2,b^2)$, thus we must have $p|\gcd(a,b)$. – asatzhh Aug 5 '13 at 12:47

$$ap+bq=1$$

implies (squaring and multiplying by $ap$) $$a^2ap^3+2p^2qa^2b+b^2apq^2=ap$$

and (similarly)

$$a^2bp^2q+2pq^2ab^2+b^2bq^3=bq$$

$$1=a^2(bp^2q+ap^3+2p^2qb)+b^2(2pq^2a+bq^3+apq^2)$$
For any two integers $x,y$, $gcd(x,y)=1$ iff there exist $r,s\in \mathbb Z$ such that $rx+sy=1$.
So now, suppose $gcd(a,b)=1$ and write $ra+sb=1$. Square both sides and rearrange terms appropriately, and this proves that $gcd(a^2,b)=1$, and also that $gcd(a,b^2)=1$. This is presumably what you've done. So, for all $a,b\in \mathbb Z$, $gcd(a,b)=1$ implies $gcd(a^2,b)=gcd(a,b^2)=1$. Apply again and you get $gcd(a^2,b^2)=1$.