'This may be completely dumb, but is it always possible to solve x/y = a/b + c/d, where all algebraic quantities are positive integers, for all x and y?' 'Isn't the general solution x = k(ad + bc), y = kbd ??' No. We get to x/y = (ad+bc)/bd. Let bd=ky, then ad+bc=kx But as x and y are pre-determined, we must select a k and therefore some partition of bd such that bd=ky, and with these values of k,b and d, then find an a and c (assuming that they exist for the chosen b,d and k values, which does not have a deterministic solution - unlike the simple rearrangement into x/y - a/b = c/d, where picking any valid a,b combo yields the values of c and d immediately. Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun