IMO 1967 LL MON33

Origin: MON

Problem

In what case does the system x + y + mz = a, x + my + z = b, mx + y + z = c, have a solution? Find the conditions under which the unique solution of the above system is an arithmetic progression.

Solution

If m ̸\in{−2, 1}, the system has the unique solution x = b + a −(1 + m)c (2 + m)(1 −m) , y = a + c −(1 + m)b (2 + m)(1 −m) , z = b + c −(1 + m)a (2 + m)(1 −m) . The numbers x, y, z form an arithmetic progression if and only if a, b, c do so. For m = 1 the system has a solution if and only if a = b = c, while for m = −2 it has a solution if and only if a + b + c = 0. In both these cases it has infinitely many solutions.