Here's what I have; The problem (4/n=1/a+1/b+1/c for all n>1 with a,b,c integers) has easy solutions for n=0,2,3mod4. If n=4k+1, then consider; 4/(4k+1) - 1/a - 1/b = 1/c Rearranging gives; c=(4k+1)ab/[4ab-(4k+1)(a+b)] wlog, let a=x(4k+1), as this is required for c to be an integer: Simplifying; c=(4k+1)xb/[4bx-4kx-x-b] Again, b=yx; c=(4k+1)xy/[4xy-4k-1-y] For c to be an integer, [1] 4xy-y=z(4k+1), and then; [2] c=xy/(z-1) So solutions to ESC for n=4k+1 are given by [1] and [2]. But here is where I get stuck. 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