Find all postive integer solutions to the equation x^yy^x=x+y.
Leming and Jer found the same two solutions, with programming, but they gave no proof of existence of another one (or anothers).
My thought is :
x^y  y^x = x+ y
x^y  x = y^x + y
x^y  x = y^x  y + y + y
x [x^(y1)  1] = y[y^(x1)  1] + 2y
The two expressions between parentheses are, respectively, divisible by (x1) and by (y1).
So, [x^(y1)  1] = k1.(x1) for some integer value of k1.
And [y^(x1)  1] = k2.(y1) for some integer value of k2.
Then :
x.k1(x1) = y.k2.(y1) + 2y
Or :
k1.x^2  k1.x  [k2.y(y1) + 2y) = 0
This is an second degree equation in x, so it has ONLY TWO roots, that are those already found.
So, I think that there are no more solutions to be searched.
Am I wrong ?

Posted by ARLEKIM
on 20050306 13:35:26 