Given positive integers a and b such that $a\mid b^2, \:b^2\mid a^3, \:a^3\mid b^4,\: b^4\mid a^5$..., prove $a=b.$ 
Given positive integers a and b such that $a\mid b^2, \:b^2\mid a^3, \:a^3\mid b^4,\: b^4\mid a^5$..., prove $a=b.$

I was able to show that a and b have the same primes in their factorizations, but I'm not sure how to show that the exponents on these primes are equivalent for all primes in their factorization. I tried contradiction but got stuck.

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*Let $v_{p}(a)$ denote the highest power of $p$ in $a$. For example if $v_{p}(a) =k$ then $p^{k}\mid a$ and $p^{k+1} \nmid a$.

*$a^{4n+1} \mid b^{4n+2} \implies v_{p}(a) \leq \frac{4n+2}{4n+1}v_{p}(b)$ and $n \to \infty$ gives $v_{p}(a) \leq v_{p}(b)$. Now try showing the reverse inequality.
