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.

Assume by contradiction $a \neq b$. Then either $a<b$ or $a>b$.
Case 1: If $a <b$ then as $b^{2n}|a^{2n+1}$ we have $b^{2n} \leq a^{2n+1}$.
Therefore
$$(\frac{b}{a})^{2n} \leq a$$
But since $\frac{b}{a} >1$ we also have
$$\lim_n (\frac{b}{a})^{2n} = \infty$$
This is a contradiction.
Case 2: If $b <a$ then as $a^{2n-1}|b^{2n}$ we have $a^{2n-1} \leq b^{2n}$.
Therefore
$$(\frac{a}{b})^{2n-1} \leq b$$
But since $\frac{a}{b} >1$ we also have
$$\lim_n (\frac{a}{b})^{2n-1} = \infty$$
This is a contradiction.
