Emmy Dillon

2022-06-26

Number theory problem, fractions and gcd
The problem says:
If $a$ and $b$ are positive integers such that $\frac{a+1}{b}+\frac{b+1}{a}$ is an integer, then show that $\sqrt{a+b}\ge gcd\left(a,b\right)$
Adding $\frac{2ab}{ab}$ to $\frac{a+1}{b}+\frac{b+1}{a}$ yields that ab divides (a+b+1)(a+b), but I haven´t been able to continue from there.

Carmelo Payne

Expert

If $gcd\left(a,b\right)=1$, then clearly $\sqrt{a+b}\ge gcd\left(a,b\right)$. Now assume $gcd\left(a,b\right)=d>1$. Then if $\frac{a+1}{b}+\frac{b+1}{a}=n$, then by your simplification, we have
$\left(a+b+1\right)\left(a+b\right)=\left(n+2\right)ab$
Note that ${d}^{2}$ divides the righthand side, so it must also divide the lefthand side. Since $d$ divides $a$ and $d$ divides $b$, $d$ must divide $a+b$. Since $a+b+1$ and $a+b$ are coprime, and since $d$ divides $a+b$, we must also have ${d}^{2}$ dividing $a+b$. In particular, $a+b\ge {d}^{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\sqrt{a+b}\ge d$

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