(x1, y1, z1) and (x2, y2, z2) are triples of real numbers such that for every pair of integers (m, n) at least one of x1m + y1n + z1, x2m + y2n + z2 is an even integer. Prove that one of the triples consists of three integers.
Solution
The basic idea is that if a real k is not an even integer, then for no real x are both x+k and 2x+k even integers. For if they were, then so would be 2(x+k)-(2x+k) = k.
Putting m = n = 0, we conclude that either z1 or z2 is an even integer. wlog we may assume z2 is an even integer. There are two cases. Suppose first that z1 is not an even integer. Then the argument above shows that at least one of x1 + z1 and 2x1 + z1 is not an even integer. Hence either x2 + z2 or 2x2 + z2 is an even integer and so x2 is an integer. A similar argument shows that y2 is an even integer and we are home.
The second case is that z1 is also an even integer. Now one of y1 + z1 and y2 + z2 is an even integer, so one of y1 and y2 is an even integer. Suppose both are. Then since one of x1 + (y1 + z1) and x2 + (y2 + z2) is an even integer, it follows that one of x1 and x2 is an even integer and we are done. So suppose that just one of y1, y2 is an even integer. wlog y2 is an even integer and y1 is not.
But that means that y1 + z1 is not an even integer. So it follows that one of x1 + (y1 + z1) and 2x1 + (y1 + z1) is not an even integer. Hence one of x2 + (y2 + z2) and 2x2 + (y2 + z2) is an even integer. So one of x2, 2x2 is an even integer, so x2 is an integer and we are home.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03