a+b divides ab+1 and a-b divides ab -1

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a+b divides ab+1 and a-b divides ab -1

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Source: 2013 Cuba 2.8

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#1 h Sep 20, 2024, 9:45 PM

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Prove that there are infinitely many pairs $(a, b)$ of positive integers with the following properties:
$\bullet$ $a+b$ divides $ab+1$ ,
$\bullet$ $a-b$ divides $ab -1$ ,
$\bullet$ $b > 2$ and $a > b\sqrt3 - 1$ .

This post has been edited 1 time. Last edited by parmenides51, Sep 20, 2024, 9:51 PM

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#2 h Sep 20, 2024, 10:57 PM

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Note that $(a,b)$ are coprime. Note that $b\equiv -a\pmod{a+b}$ yields $a+b\mid b^2-1$ . Likewise, $a\equiv b\pmod{a-b}$ gives $a-b\mid b^2-1$ . Equipped with this, let's choose $a,b$ such that (a) both $a+b$ and $a-b$ are even, and (b) $a+b,a-b\mid b^2-1$ . For this, it suffices to ensure $a^2-3b^2=-2$ has infinitely many odd solutions with $b>2$ . Notice that if $a^2=3b^2-2$ then $a>b\sqrt{3}-1$ necessarily, as otherwise $3b^2-2b\sqrt{3}+1>3b^2-2$ , not possible for $b$ large.

Any such odd solution would necessarily imply $a,b$ are coprime. Furthermore, $ab+1\equiv b^2-1\equiv \frac{a^2-b^2}{2} = (a+b)\frac{a-b}{2}\equiv 0\pmod{a+b}$ , and $a-b\mid ab-1$ is established similarly. The infinitude of the solutions of the above equation is well-known, see e.g. here.

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