Let a, b, c be positive integers such that a and b are relatively prime and c is relatively prime to a or b. Show that there are infinitely many solutions to ma + nb = kc, where m, n, k are distinct positive integers.
Solution
Many thanks to Kamil Duszenko for the following solution
If a or b = 1, then the problem is trivial (if a = 1, then just put m = kc - nb. We can obviously get m, n, k distinct and positive for infinitely many choices of k and n). So assume a, b > 1.
Now take m = 2tb, n = 2ta, k = (tab + 1)/c. Since c is relatively prime to ab, we can find t such that tab = -1 mod c. Then all t + c, t + 2c, t + 3c, ... have the same property. So there are infinitely many t for which m, n, k are positive integers. Obviously, ma + nb = kc, and m and n are distinct. Suppose m = k. Then tb = tab + 1, so b = 1. Contradiction. Similarly, n cannot be k. So m, n, k are all distinct.
© John Scholes
jscholes@kalva.demon.co.uk
7 July 2003
Last corrected/updated 7 Jul 2003