Let $a,b$ be positive integers such that $a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots$ so on , then $a=b$?

Solution 1:

Let $p$ a prime that divides $a$ then since $a|b^2$ then $p$ divides $b$. Similarly we get that if $p$ is a prime that divides $b$ then it divides $a$ hence $a$ and $b$ has the same primes in their primary decomposition. Now if $p^\alpha$ is the factor of the primary decomposition of $a$ and $p^\beta$ is that of $b$ then

$$\alpha\le2\beta\le3\alpha\le 4\beta\le\cdots$$ then

$$\alpha\le\frac {2n}{2n-1}\beta,\quad \forall n$$ and $$\beta \le\frac {2n+1}{2n}\alpha,\quad \forall n$$ so by taking the limit $n\to\infty$ we get $\alpha=\beta$ and then $a=b$.

Solution 2:

If $a <b$ then $$\lim_n\frac{b^{2n}}{a^{2n+1}}= \infty$$

which means that from some point $b^{2n} >a^{2n+1}$, and it cannot divide it.

Same way, if $a>b$ then

$$\lim_n\frac{a^{2n-1}}{b^{2n}}= \infty$$

Solution 3:

This is a twist of another answer here in order to avoid prime factorization.

We may assume $a,b>1$. Then $$\frac{2n-1}{2n} \leq \frac{\log b}{\log a} \leq \frac{2n+1}{2n}$$ for all $n$. Taking $n \to \infty$, we get $a=b$.