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.

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