Problem: Prove that there exists a unique triple of positive integers greater than 1 such that product of any two increased by one is divisible by the third number.

I found $(2,3,7)$ to satisfy this but cant prove this. Please help.

  • 1
    $\begingroup$ Please give the contest this comes from. It will help to know what level of math is expected in the solution. Can you prove that the three integers are distinct? $\endgroup$ – Ross Millikan Oct 4 '16 at 2:39

Clearly all numbers must be pairwise coprime, let them be $a<b<c$

Notice that $abc| ab+ac+bc+1$ and so $abc\leq ab+ac+bc+1$.

Notice that if $a\geq 3$ then $abc\geq 3bc> ab+ac+bc+1$ and so $a=2$.

It follows that $b|2c+1$ and $c|2b+1$ and so $bc|2b+2c+1$ In particular $bc\leq 2b+2c+1$.

It $b\geq 4$ then $bc\geq 4c>2b+2c+1$ and so $b=3$.

Hence $a=2,b=3$. And finally we must have $c|2\times 3+1$. So $c=7$.

  • $\begingroup$ how does $abc|ab+bc+ca+1$? $\endgroup$ – user366265 Oct 4 '16 at 3:12
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    $\begingroup$ Because each of $a,b,c$ divides it, and they're coprime. $\endgroup$ – arkeet Oct 4 '16 at 3:17
  • $\begingroup$ yeah, that :) ${}$ $\endgroup$ – Jorge Fernández Hidalgo Oct 4 '16 at 3:17

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