If $a\mid b^2, b^2\mid a^3,\ldots ,a^n\mid b^{n+1},b^{n+1}\mid a^{n+2},\ldots$ then $a=b$ I'm stuck with this problem :
Let $a,b$ positive integers such that
$$a\mid b^2, b^2\mid a^3,\ldots ,a^n\mid b^{n+1},b^{n+1}\mid a^{n+2},\ldots$$
Show that $a=b$.
If were $ b > a $ then $\lim_{n \to \infty}\frac{b^n}{a^n}=0 $ choosing $\epsilon = \frac{1}{a}$ we got a contradiction, but I can't show that $b<a$ can't hold.
Any help is apreciated.

Pick a prime $p$, we must show $v_p(a)=v_p(b)$.
From the divisibility relations we have, for each $n\in \mathbb Z^+$:
$v_p(a)\leq \frac{v_p(b)2n}{2n-1}\implies v_p(a)\leq v_p(b)$.
We also have:
$v_p(b)\leq \frac{v_p(a)2n+1}{2n}\implies v_p(b)\leq v_p(a)$
