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 $a=b$ 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 $a=b$
 A: 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.
A: 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?
A: 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
Fractions whose powers have such a common denominator are called almost integral. It is clear that integral elements are almost integral. By above  the converse is true in Noetherian domains.
A: 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,
