Solution May 25, 2007

Problem

Given natural numbers $a\,$, $b\,$ and $c\,$ that are pairwise distinct and satisfy $a | b+c+bc,\, b | c+a+ca,\, c | a+b+ab\,$,

prove that at least one of the numbers $a\,$, $b\,$, $c\,$ is not prime.

Solution

Assume that a,b,c are distinct primes. Further assume without loss of generality that a < b < c.

Note that $b+c+bc = (b+1)(c+1)-1\,$, so $a|(a+1)(b+1)(c+1)-1\,$ and similar for $b\,$ and $c\,$. Also, $a,\, b,\, c$ are pairwise coprime, so we have

$abc | (a+1)(b+1)(c+1)-1.\,$

Let

$n := \frac{(a+1)(b+1)(c+1)-1}{abc} = 1 + \frac 1a + \frac 1b + \frac 1c + \frac 1{ab} + \frac 1{ca} + \frac 1{bc}.\,$.

$n\,$ is an integer. Obviously, $n>1\,$. The right side is a strictly decreasing function in all three variables. Let's check a few cases:

• if $a = 2\,$ and $b = 3\,$, the right side becomes $2 + \frac1{c} + \frac1{2c} + \frac1{3c} = 2 + \frac{11}{6c}\,$ which is never an integer for any $c\,$
• if $a = 2\,$, $b = 5\,$, the right side is not an integer for $c=7\,$ and less than $2\,$ for $c\geq 11\,$, so it's not an integer then either.
• if $a = 2\,$, $b=7\,$ then $c\geq 11\,$ and again, $n<2\,$.
• if $a \geq 3\,$, then $b\geq 5\,$ and $c\geq 7\,$ and, (surprise!) $n<2\,$.

This shows that $n\,$ can never be an integer, in contradiction with our assumptions, completing the proof.