Jon Perry (perry@globalnet.co.uk) wrote:

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:

If you mean that x must be an integer, this isn't true. For example, consider 4/5 = 1/2 + 1/5 + 1/10: k=1, a=2, b=5, c=10, x=2/5.

Simplifying;

c=(4k+1)xb/[4bx-4kx-x-b]

Again, b=yx;

If you mean that y must be an integer, this isn't true. For example, consider 4/29 = 1/87 + 1/8 + 1/696: k=7, a=87, b=8, c=696, x=3, y=8/3.

c=(4k+1)xy/[4xy-4k-1-y]

For c to be an integer, [1] 4xy-y=z(4k+1), and then;

If you mean that z must be an integer, this isn't true. For example, consider 4/5 = 1/5 + 1/2 + 1/10: k=1, a=5, b=2, c=10, x=1, y=2, z=6/5. Dean Hickerson dean@math.ucdavis.edu