Solution May 25, 2007


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.


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.\,


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:

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