Find all positive integers m, n such that m + n2 + d3 = mnd, where d is the greatest common divisor of m and n.
Solution
Answer: (m, n) = (4, 2), (4, 6), (5, 2) or (5, 3).
Put m = Md, n = Nd, so that M and N are coprime. Then: M + N2d + d2 = MNd2. So d must divide M. Put M = M'd, then M' + N2 + d = M'Nd2.
So M' = (N2 + d)/(Nd2 - 1). Hence M'd2 = N + (N+d3)/(Nd2 - 1). So (N + d3)/(Nd2 - 1) is an integer. If d = 1, (N+1)/(N-1) can only be an integer for N = 2 or 3. That gives the solutions (m, n) = (5, 2) and (5, 3). So assume d > 1. Since (N + d3)/(Nd2 - 1) is an integer and positive, we must have N + d3 >= Nd2 - 1, so N <= (d3 + 1)/(d2 - 1). If d = 2, then N <= 3. Then N = 1 gives the solution (m, n) = (4, 2), N = 2 gives (N + d3)/(Nd2 - 1) non-integral and hence no solution, N = 3 gives the solution (m, n) = (4, 6).
So assume d > 2. Then (d3 + 1)/(d2 - 1) < d + 1. Hence N ≤ d. Hence M' = (N2 + d)/(Nd2 - 1) <= (d2 + d)/(Nd2 - 1) <= (d2 + d)/(d2 - 1), since N ≥ 1 and hence M' < 2 (since d > 2). So M' = 1. So N is a root of the quadratic N2 - d2N + d+1 = 0. But d4 - 4(d+1) > d4 - 4d2 + 1 = (d2 - 1)2 and < d4, so d4 - 4(d+1) cannot be a square and hence N cannot be integral. So there are no solutions with d > 2.
© John Scholes
jscholes@kalva.demon.co.uk
8 Oct 2002