If $a, b$ are positive integers such that $a|b^2, b^2|a^3,....$ then $a=b$. The following is a well known problem 
If $a|b^2; b^2|a^3; a^3|b^4; ....$ then $a=b$.
I started with $a$ is a prime $p$ and the conclusion immediately holds. 
So I was comparing powers of primes but can't get anything useful.
 Please help.
 A: Hint $\quad\ \forall n\in\mathbb N: \ \color{#c00}a\:\!\left(\dfrac{a}b\right)^{2n}\!\!\in\mathbb Z\,\Rightarrow \color{#c00}a\,$ is a denominator for unbounded powers of $\,\dfrac{a}b\,$ so $\,\dfrac{a}b\in\Bbb Z$
Likewise $\forall n\in\mathbb N:\ b\:\!\left(\dfrac{b}a\right)^{2n+1}\!\!\in \mathbb Z\ \ \Rightarrow\ \dfrac{b}a\in \Bbb Z.\ $ Thus $\ b\mid a\mid b\ $ so $\,a = b.$
Remark $ $ This generalizes to any Noetherian integrally-closed domain $D,\,$ e.g. any PID, since if unbounded powers of a fraction $f$ over $D$ have a common denominator then $\,f\in D\,$ (Proof)
A: Write:
$$a = \prod_1^m p_i ^{t_i}$$
$$b=\prod_1^m p_i ^{s_i}$$
where $t_i \ge 0$ and $s_i \ge 0$. 
The given says that given any $i$, 
$$\forall n\in \mathbb N: \begin{cases} t_i \le \frac{2n}{2n-1} s_i \\ s_i \le \frac{2n+1}{2n} t_i\end{cases}$$
Taking $n \to \infty$, we get $t_i \le s_i \le t_i$, i.e. $s_i = t_i$.
A: Hint:  that is a good start.  Now can you prove it for $a=p^k$ with any prime $p$?  $b$ has to again be a power of $p$.  After that, claim that it is multiplicative, so when you factor $a$ you get the two cases you have proved.
