a problem

[nfstiham]

x^y=y^x where x is inequal to y.solve it.how many solution does it have?

[Hasib]

x, y are positive real number?

[nfstiham]

just find me another answer except (2,4).it can be anything.

[Babai]

There is a good way to prove this.Actually I am proving this from a different aspect.

You can prove by induction that $x=(1+\frac{1}{n})^n$ and $y=(1+\frac{1}{n})^{n+1}$ or the converse for $x$ is not equal to $y$ except $(x,y)=(1,1);(0,0)$

So,there exist many real solutions.

[*Mahi*]

Let $x=ny$ (you will know why I take this very soon)
Then,$x^{nx}=(nx)^x$
Or,$x^{(n-1)x}.x^x=n^x.x^x$
Thus,$x^{(n-1)x}=n^x$
Or,$x^{n-1}=n$
Or,$x=n^{\frac {n}{n-1}}$
And $y=n^{ \frac 1 {n-1}}$
So,we get a solution for every real number $n$.

[Masum]

x, y are positive real number?

Of-course not.Since it is posted in number Theory forum,it is assumed to be a diophantine equation.Otherwise,it would be posted in Algebra forum.

[Avik Roy]

Nice Problem. Giving a hint only-
Try with prime power factorization of $x$ and $y$. That'll lead you to some handy equations

[Masum]

Yes,note that $x$ and $y$ must have the same prime factors(why?)

[tarek like math]

$x^y=y^x$. first y should be $y=x^a$ then $x^a=ax$ for that $a=x^(a-1)$ and x should be $x=x^a/a$. so $x^(x^a)=(x^a)^(x^a/a)=x^(x^a)$
an equation can come as such $a=x^(a-1) or a=(x^a/a)^(a-1) or a=x^(a^2-a)/a^(a-1) or a=x^(a^2-a)/x^(a-1)(a-1)$ or $a=x^(a^2-a-a^2+2a-1) or a=x^(a-1)$.
possible integers can be found by solution of this $x^(a^2-a)=a^a$.

[tarek like math]

Of course both x and y have same prime factors