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Let $a,b \in {\mathbb{Z_+}}$ such that $$a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$$

Prove $a=b$

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    $\begingroup$ Show that if $a<b$ then $b^n>a^{n+1}$ for large values of $n$. $\endgroup$ – Thomas Andrews Mar 23 '12 at 12:22
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consider the prime factorisations of $a = \prod_p p^{\nu_p(a)}$ and $b = \prod_p p^{\nu_p(b)}$. Your assumptions yield

  • $\nu_p(a) \le \nu_p(b^2) = 2\nu_p(b)$ for each $p$
  • $3\nu_p(b) \le 4\nu_p(a)$
  • $\ldots$
  • $(4n+1)\nu_p(a) \le (4n+2)\nu_p(b)$ for each $p$, $n$
  • $(4n+3)\nu_p(b) \le (4n + 4)\nu_p(a)$, each $p$, $n$

So we have for each $p$, $n$ $$ \frac{4n + 3}{4n+4} \cdot \nu_p(b) \le \nu_p(a) \le \frac{4n+2}{4n+1}\nu_p(b) $$ letting $n\to \infty$ yields $\nu_p(a) = \nu_p(b)$ for each $p$, so $a = b$.

AB,

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If $a > b$ then $\frac{a}{b}>1$ and hence there is an $n$ such that $\left(\frac{a}{b}\right)^n > b$, thus $a^n > b^{n+1}$. This contradicts $a^{4k+1} | b^{4k+2}$. The case $a<b$ works in just the same way.

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  • $\begingroup$ I like that this solution makes it clear that divisibility is a red herring---the problem would work just as well with the much weaker hypothesis that $a \leq b^2$, $b^3 \leq a^4$, etc., for real values of $a$ and $b$. $\endgroup$ – Dan Petersen Mar 23 '12 at 15:08
  • $\begingroup$ @Dan Divisibility isn't necessarily a red herring, since the proof I gave employs divisibility, and it works in rings where the (archimedean) order-based proofs do not apply, e.g. the Gaussian integers $\mathbb Z[i]$ or the ring of integers of any (complex) algebraic number field. $\endgroup$ – Bill Dubuque Mar 24 '12 at 21:50
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Let $p$ be a prime, and suppose $p^7$ divides $a$, but $p^8$ doesn't. Then from $a^9\mid b^{10}$ you can deduce $p^7\mid b$ (why?). Now if $p^8\mid b$, then from $b^{11}\mid a^{12}$ you would get $p^8\mid a$ (why?), contradiction. So same power of $p$ divides $a$ and $b$.

Can you see how to generalize this argument?

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Hint $\rm\ \ \forall n\in\mathbb N:\ \ a\:\!\left(\dfrac{a}b\right)^{4n+3}\!\in\mathbb Z,\:\ b\:\!\left(\dfrac{b}a\right)^{4n+1}\!\in \mathbb Z\ \ \Rightarrow\ \dfrac{a}b,\:\dfrac{b}a\in\mathbb Z\ \ \Rightarrow\ \ a = \pm b\ \ \ $ QED

Remark $\ $ This is true far more generally. Suppose $\rm\:D\:$ is any Noetherian integrally closed domain, e.g. any PID. Suppose that $\rm\:w\:$ is a fraction over $\rm\:D\:$ such that some unbounded sequence of powers of $\rm\:w\:$ has a common denominator $\rm\:0 \ne d\in D,\:$ i.e. $\rm\:d\!\:w^{n_i}\in D\:$ for all $\rm\:n_i.\:$ Then $\rm\:w\in D.$

Proof $\ $ By ACC the sequence of ideals $\rm (d, dw^{n_1}, dw^{n_2},\ldots)$ eventually stabilizes, which implies that for some $\rm\:k\:$ we have $\rm\: dw^{n_k}\in (dw^{n_{k-1}},\ldots, dw^{n_1}, d),\:$ which implies

$$\rm d\: w^{n_k} + c_{n_{k-1}} d\: w^{n_{k-1}} +\:\! \cdots +\: c_{n_1} d\: w^{n_1} + d\: =\: 0$$

Cancelling $\rm\:d\:$ yields $\rm\:w\:$ is integral over $\rm\:D,\:$ hence $\rm\:w\in D,\:$ since $\rm\:D\:$ is integrally closed. $\ $ QED

Elements whose powers have such a common denominator are called almost integral. It is clear that integral elements are almost integral. The above shows that the converse holds true in Noetherian domains.

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