For what positive integers m, n can we find positive integers a, b, c such that a + mb = n and a + b = mc. Show that there is at most one such solution for each m, n.
Solution
We fix c and solve for a and b.
We have n = a + mb = a + m(mc - a)= mc+1 - a(m-1). Hence (n-1)/(m-1) = (mc+1-1)/(m-1) - a. So a = (mc+1-1)/(m-1) - (n-1)/(m-1). Note that (mc+1-1)/(m-1) is an integer, so a is an integer iff n-1 is a multiple of m-1. Also a is positive iff mc+1 > n.
b = mc - a = (n-1)/(m-1) - (mc-1)/(m-1). So b is also an integer iff n-1 is a multiple of m-1, and is positive iff n > mc.
If n is a power of m, then we cannot find an integer c such that mc+1 > n > mc, so there are no solutions. Otherwise, there is a unique c such that mc+1 > n > mc.
Thus we can find positive integers a, b, c iff m > 1, n-1 is a multiple of m-1, n is not a power of m and the solution is then unique.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03