Prove that $ab$ is a cube of a positive integer.

Question

This is a question from Advanced Math Examination of Vietnam:

Let $a, b, c$ be the three positive integers such that $c+\frac{1}{b}=a+\frac{b}{a}$. Prove that $ab$ is a cube of a positive integer.

First solution I thought about is form the hypothesis to a more similar one by move $\frac{1}{b}$ to the other side to make an inverse expression.

I have tried to form $a.b=c^{3}$ using the hypothesis $c+\frac{1}{b}=a+\frac{b}{a}$ but I can't find a good way to simplify the equation. It's too complicated.

Answer 1

Note that

$$\frac{b}{a}-\frac{1}{b} \in \mathbb Z$$ and hence $ab|b^2-a$. This implies that $b|a$ and $a|b^2$.

This implies that there exists integers $d,e$ such that $$a=bd \\ b=de \Rightarrow a=d^2e$$

Simplification suggested by Will Jagy

Therefore, $ab|b^2-a$ means $d^3e^2|d^2e(e-1)$ and hence $de|e-1$. This implies $e|e-1$ from where it follows that $e=1$.

Then: $ab=d^3$.

Note The simplification suggested by Will Jagy actualy proves the following much stronger claim: If $\frac{b}{a}-\frac{1}{b} \in \mathbb Z$ then $a=b^2$.

It is trivially to see that the converse is also true.