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The following is a well known problem

If $a|b^2; b^2|a^3; a^3|b^4; ....$ then $a=b$.

I started with $a$ is a prime $p$ and the conclusion immediately holds. So I was comparing powers of primes but can't get anything useful. Please help.

Martin Sleziak
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3 Answers3

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Hint $\quad\ \forall n\in\mathbb N: \ \color{#c00}a\:\!\left(\dfrac{a}b\right)^{2n}\!\!\in\mathbb Z\,\Rightarrow \color{#c00}a\,$ is a denominator for unbounded powers of $\,\dfrac{a}b\,$ so $\,\dfrac{a}b\in\Bbb Z$

Likewise $\forall n\in\mathbb N:\ b\:\!\left(\dfrac{b}a\right)^{2n+1}\!\!\in \mathbb Z\ \ \Rightarrow\ \dfrac{b}a\in \Bbb Z.\ $ Thus $\ b\mid a\mid b\ $ so $\,a = b.$

Remark $ $ This generalizes to any Noetherian integrally-closed domain $D,\,$ e.g. any PID, since if unbounded powers of a fraction $f$ over $D$ have a common denominator then $\,f\in D\,$ (Proof)

Bill Dubuque
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Write:

$$a = \prod_1^m p_i ^{t_i}$$

$$b=\prod_1^m p_i ^{s_i}$$

where $t_i \ge 0$ and $s_i \ge 0$.

The given says that given any $i$,

$$\forall n\in \mathbb N: \begin{cases} t_i \le \frac{2n}{2n-1} s_i \\ s_i \le \frac{2n+1}{2n} t_i\end{cases}$$

Taking $n \to \infty$, we get $t_i \le s_i \le t_i$, i.e. $s_i = t_i$.

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Hint: that is a good start. Now can you prove it for $a=p^k$ with any prime $p$? $b$ has to again be a power of $p$. After that, claim that it is multiplicative, so when you factor $a$ you get the two cases you have proved.

Harsh Kumar
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Ross Millikan
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  • I concluded that $a,b$ will have the same prime factors. Regarding their powers I am having troubles with inequalities. Can you please give a bit elaboration on that part? –  Jun 05 '17 at 02:53
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    Let $a=p^k, b=p^m$. The first tells you that $k \le 2m$. The second tells you that $2m \le 3k$. If $m \neq k$ you can find some high ones that have a ratio closer to $1$ than $\frac mk$ – Ross Millikan Jun 05 '17 at 03:03
  • Thanks a lot sir. –  Jun 05 '17 at 03:05
  • I don't get how to prove the case when $a=p^k$ Can you explain me please? – Vmimi Nov 06 '18 at 02:25
  • @Vmimi: If $a=p^k$ the second tells you that $b$ has only $p$ for a prime factor, so $b=p^q$ and $2q \le 3k$. The third says $3k \le 4q$ and so on. If $k \neq q$ there will be one of the divisions that will fail. – Ross Millikan Nov 06 '18 at 02:33
  • I still don't get it. Sorry – Vmimi Nov 07 '18 at 01:17
  • Do you see that $b=p^q?$ Then if $q \gt k,$ we just need to go high enough in the chain of divisibility to find one that fails because the ratio of the coefficients approaches $1$. For example, assume $k=100,q=101$. We are told $b^{500}|a^{501}$, but $b^{500}=p^{500\cdot 101}=p^{50500}, a^{501}=p^{501\cdot 100}=p^{50100}$ and the divisibility fails. – Ross Millikan Nov 07 '18 at 01:24