Find all postive integer solutions to the equation x^y-y^x=x+y.
(In reply to
Am I wrong ? by pcbouhid)
You wrote:
So, [x^(y-1) - 1] = k1.(x-1) for some integer value of k1.
And [y^(x-1) - 1] = k2.(y-1) for some integer value of k2.
but k1 and k2 are polynomials. For example if y=3, k1=(x+1).
My hope would be that since as you found an x^2 this somehow yields the two solutions we found and this additional polynomial give no others.
|
Posted by Jer
on 2005-03-07 17:04:42 |