If $\gcd(a,b) = p$, prime number then, what we can tell about the gcd of $(a^2,b^3)$? 
If $\gcd(a,b) = p$, prime number, then what we can tell about $\gcd(a^2,b^3)$?

No idea how to solve.

Raise the equation
$$
\frac apx+\frac bpy=1
$$
to the fourth power
$$
\frac{a^4}{p^4}x^4+4\frac{a^3}{p^3}x^3\frac bpy+6\frac{a^2}{p^2}x^2\frac{b^2}{p^2}y^2+4\frac apx\frac{b^3}{p^3}y^3+\frac{b^4}{p^4}y^4=1
$$
which is
$$
a^2p\left(\frac{a^2}{p^2}x^4+4\frac{a}{p}x^3\frac bpy+6x^2\frac{b^2}{p^2}y^2\right)+b^3\left(4\frac apxy^3+\frac{b}{p}y^4\right)=p^3
$$
Thus, $(a^2,b^3)\mid p^3$. Since $p\mid a$ and $p\mid b$, we have that $p^2\mid(a^2,b^3)$. Therefore,
$$
(a^2,b^3)\in\left\{p^2,p^3\right\}
$$
