number theory - Proving $amid b^2,,b^2mid a^3,,a^3mid b^4,ldotsimplies a=b$ - why is my approach incorrect

https://math.stackexchange.com/questions/704048/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?

Answer

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.