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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.

Mermaid Aine
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1 Answers1

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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.

N. S.
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    It's not immediately obvious that $ab|b^2-a$ $\implies$ $b|a$ and $a|b^2$. Can you shed a light why? Thanks. – Alexey Burdin Jul 21 '20 at 04:13
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    @Alexey Burdin If $ab|b^2-a$ $\implies kab=b^2-a\implies a(kb+1)=b^2\implies a|b^2...$ – AsdrubalBeltran Jul 21 '20 at 04:32
  • @AlexeyBurdin I don't understand how you get $e=1$ – Mermaid Aine Jul 21 '20 at 05:24
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    @MermaidAine $ab|b^2-a$ implies $a|b^2-a$. Since $a|-a$ this means $a|b^2$. Same way $b|b^2-a$ and $b|b^2$ implies $b|a$/ – N. S. Jul 21 '20 at 05:38
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    @MermaidAine $$1=e(d^3e+1-cd)$$ – N. S. Jul 21 '20 at 05:38
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    NS, without dragging in the integer $c$ again, your relation $ab| b^2-a$ with $a=d^2 e$ and $b=de$ gives $d^3 e^2 | d^2 e (e-1),$ or $de|e-1.$ – Will Jagy Jul 21 '20 at 17:50
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    @WillJagy Nice observation, see the edit. – N. S. Jul 21 '20 at 19:26
  • good. I noticed the infinite family $a=b^2$ last night but was not sure those were all. – Will Jagy Jul 21 '20 at 19:31
  • @WillJagy I think that the proof can be simplified a bit but it is probably not worth: $\frac{b}{a}-\frac{1}{b} \in \mathbb Z \Rightarrow b|a \Rightarrow a=bd \Rightarrow \frac{1}{d}-\frac{1}{b} \in \mathbb Z$. Now the difference is strictly between $-1$ and $1$. – N. S. Jul 21 '20 at 19:33