Using Bezout's identity I have shown


but what should be the next step? Thanks.

  • 4
    $\begingroup$ HINT: for any prime $p$ , $p|\gcd(a^2,b^2)$, then $p^2|\gcd(a^2,b^2)$, thus we must have $p|\gcd(a,b)$. $\endgroup$ – asatzhh Aug 5 '13 at 12:47

Just to follow your idea with Bezout's.


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

and (similarly)


Adding them



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


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