Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ so on , then $a=b$? Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ that is $a^{2n-1}\mid b^{2n} ; b^{2n}\mid a^{2n+1} , \forall n \in \mathbb Z^+$ , then is it true that $a=b$ ?
 A: Let $p$ a prime that divides $a$ then since $a|b^2$ then $p$ divides $b$. Similarly we get that if $p$ is a prime that divides $b$ then it divides $a$ hence $a$ and $b$ has the same primes in their primary decomposition. Now if $p^\alpha$ is the factor of the primary decomposition of $a$ and $p^\beta$ is that of $b$ then
$$\alpha\le2\beta\le3\alpha\le 4\beta\le\cdots$$ then
$$\alpha\le\frac {2n}{2n-1}\beta,\quad \forall n$$
and
$$\beta \le\frac {2n+1}{2n}\alpha,\quad \forall n$$
so by taking the limit $n\to\infty$ we get $\alpha=\beta$ and then $a=b$.
A: If $a <b$ then $$\lim_n\frac{b^{2n}}{a^{2n+1}}= \infty$$
which means that from some point $b^{2n} >a^{2n+1}$, and it cannot divide it.
Same way, if $a>b$ then
$$\lim_n\frac{a^{2n-1}}{b^{2n}}= \infty$$
A: This is a twist of another answer here in order to avoid prime factorization.
We may assume $a,b>1$. Then
$$\frac{2n-1}{2n} \leq \frac{\log b}{\log a} \leq \frac{2n+1}{2n}$$
for all $n$. Taking $n \to \infty$, we get $a=b$.
A: Under the given condition, it is clear that the prime factors of $a$ and $b$ are the same. 
Without loss of generality, you can assume that $a = p^i$ and $b=p^j$ where $p$ is a prime factor and $i,j\in \mathbb N$ (for if $a_k$ and $b_k$ are of this form and fulfill the given condition, it is clear that $a$ = product of $a_k$ and $b$ = product of $b_k$ fulfills this condition too, and conversely etc.).
There holds $p^i|p^{2j}$, $p^{2j}|p^{3i}$ etc. 
Hence $i \leq  2j \leq 3i \leq 4j \cdots \leq n i \leq (n+1) j\leq (n+2) i$ etc. Taking $n$ large shows that $i=j$.
In conclusion, yes $a = b$.
