# Proving $a\mid b^2,\,b^2\mid a^3,\,a^3\mid b^4,\ldots\implies a=b$ - why is my approach incorrect

Theory Number Problems After I saw that post i wanted to solve the first one which is $a\mid b^2,b^2\mid a^3,a^3\mid b^4,b^4\mid a^5\cdots$ Prove that $a=b$

Now i started by proving that $a$ and $b$ have same prime divisors,proving that is trivial,after doing so I checked do those factors have to have the same power.I tried with $$a=p^2,b=p^3$$ I noticed it exactly goes for 3 terms,or that $a^3\mid b^4$ doing $$a=p^{n-1},b=p^n\\p^{n-1}\mid p^{2n}\\p^{2n}\mid p^{3n-3}\\p^{3n-3}\mid p^{4n}\\p^{4n}\mid p^{5n-5}\\\cdots\\p^{n(n-1)}\mid p^{n(n-1)}\\p^{n(n-1)}\mid p^{n(n+1)}\\p^{n(n+1)}\not\mid p^{(n-1)(n+1)}$$ Now yeah I guess that would be a proof,but wouldn't setting $n=n+1$ infinitely many times make every term dividable?

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For every prime that divides $\;a\;$, look at the maximal power of it that still divides it and show the same power divides $\;b\;$...and the other way around, of course. –  DonAntonio Mar 8 '14 at 13:48
What does "setting $n=n+1$ infinitely many times" mean? –  Andres Caicedo Aug 10 '14 at 17:20
See these questions: this and this. –  user26486 May 26 at 4:05

Hint $\ b(b/a),\,b(b/a)^3,b(b/a)^5\ldots\,$ are all integers, i.e. $\,b\,$ is a common denominator for $\rm\color{#c00}{unbounded}$ powers of $\,b/a,\,$ hence $\,b/a,$ is an integer (prove this!), therefore $\,a\mid b.\,$ Similarly $\,b\mid a.\,$
Further hint:  let $\, r = \frac{b}a = \frac{c}d,\ (c,d)=1.\,$ Then $\,br^k = n\in\Bbb Z\,\Rightarrow\, bc^k = n d^k\,$ so $\,d^k\mid b c^k\Rightarrow\, d^k\mid b\,$ by Euclid's Lemma and $\,(c,d)=1.\,$ $\,d^k\mid b\,$ for $\,\rm\color{#c00}{unbounded}$ $\,k\,$ $\,\Rightarrow\,d=\pm1,\,$ so $\,r = \frac{c}d = \pm c\in\Bbb Z.$
Remark $\$ The above property, that unbounded powers of proper fractions cannot have a common denominator (such as $b$ above), is a special property of $\,\Bbb Z\,$ that needn't be true in general domains. When it holds true, a domain is called completely integrally closed.