If $b^2\mid a^2$, then prove $b\mid a^2$.

Question

If $b^2\mid a^2$, then prove $b\mid a^2$.

To prove, need consider the fact that if $a$ is odd, then $a^2$ too is. Let, for a suitable integer $k$, $a=2k+1$, then $a^2= 4k^2+4k+1$.
So, if $b^2\mid a^2,$ then both ($a,b$) have the same parity.

There are two cases, based on the parity of $a,b$ being odd or even.
(i) If both odd: let for suitable integers $k,m$, $a=2k+1, b=2m+1$; $a^2= 4k^2 + 4k+1, b^2=4m^2+4m+1$.
=> $(4m^2 + 4m+1)\mid (4k^2 + 4k+1)$
Unable to show algebraic proof further for the next step.

(ii) if both even: it is obvious that if $b^2$ divides $a^2$, then $b\mid a^2$. but cannot show the algebraic proof.

Answer 1

$b^2\mid a^2 \implies a^2=kb^2\implies a=b\sqrt k$. Since $a$ is an integer, we can say that $k$ is perfect square. So $k=c^2\implies a^2=c^2b^2\implies a=cb$. Hence, $$b\mid a\implies b\mid a^2$$

Answer 2

Lemma If $k\mid m$ and $m\mid n$, then $k\mid n$.

Proof Indeed, since $k\mid m$, $m=kj$ for some integer $j$. Likewise $n=ml$ for some integer $l$ whence $n=k(jl)$ as desired.

For your problem observe that $b\mid b^2$ and $b^2\mid a^2$, so...