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 $a=b$

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,

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.

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?