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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 at 13:48
    
What does "setting $n=n+1$ infinitely many times" mean? –  Andres Caicedo Aug 10 at 17:20

1 Answer 1

Hint $\ b(b/a),\,b(b/a)^2,\,b(b/a)^3,\ldots\,$ are all integers, i.e. $\,b\,$ is a common denominator for all powers of $\,b/a,\,$ hence $\,b/a,$ is an integer (prove this!), therefore $\,b\mid a.\,$ Also $\,a\mid b\,$ by symmetry.

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.

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