X^y=y^x proof

Just take x=y and you have infinitely many solutions

So the best you can hope for is to find a proof for your assertion under the additional constraint that x and y be distinct.

 

Try using the prime factorization of x and y for a start

 

I think on this proof:

consider y > x

[tex]x^y = y^x ==> y = kx ==> x^y = (kx)^x ==> x^{y-x} = k^x ==> x^{x(k-1)} = k^x ==> x^{k-1} = k[/tex]

as [tex]x^{k-1} = k[/tex] is true for x > k if and only if k = 1, and k = 1 ==> x=y, this is a contradiction with our initial consideration that y > x, then k >= x

by [tex]x^{k-1} = k[/tex] it is easy to see that k and x have exactly the same prime numbers as factors.

proof: k >= x, supose that there is one factor in k not in x, so lets write [tex] k = w*x^n ==> x^{k-1 -n} = w[/tex] that is true if and only if w = 1 and n = k - 1

as x and k have exactly the same factors [tex]x^{k-1} = k[/tex] is true if and only if x=k ==> [tex]x^{x-1} = x ==> x = 2[/tex]

x = k = 2, as y = kx, then y = 4 and the proof is finish