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Let $a$ and $b$ be positive integers such that $a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5, \cdots $ Prove that $a = b$.

My way is as follows: Let $A=v_p(a), B=v_p(b)$ be the exact power of a prime $p$ dividing $a$ and $b$ respectively. Then by assumption, $(2n-1)A \leq 2nB, n=1,2,3,\cdots.$ Hence $A \leq (2n/(2n-1))B$. Taking limit as $n$ tends to infinite yields $A \leq B.$ Likewise, $2nB \leq (2n+1)A, n=1,2,3,\cdots,$ by a similar argument, we have $B \leq A.$ So that $A=B$ and we are done. But I am rather unsatisfied with the proof because it involves limit. I wanna ask if there is any other alternative proof, thank in advances.

Minori Chihara
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Hint $ $ The odd terms show that for $\,c = b/a\,$ we have $\, b c^{2n+1}\!\in \Bbb Z\,$ for all $\,n\in \Bbb N,\,$ hence $\,c\in\Bbb Z\,$ (else, by Euclid's Lemma, $b$ is divisible by unbounded powers of $c$'s reduced denominator $> 1).\,$ Hence $\,c\in\Bbb Z\,\Rightarrow\,a\mid b.\,$ Similarly $\,b\mid a\,$ from the even terms.

Remark $\ $ Domains are called completely integrally closed when they satisfy the above property, i.e. unbounded powers of proper fractions cannot have a common denominator (such as $b$ above).

Bill Dubuque
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