Solve the equation $x^{x+y}=y^{y-x}$ over natural numbers 
Solve the equation $x^{x+y}=y^{y-x} \tag 1$ where $x,y \in \mathbb{N}$


$x = 1,y = 1$ is a solution, now suppose $x \ne 1, y \ne 1$.
Obviously $x, y \ne 0$ and $x, y$ have same prime divisors. 
Because $x+y \gt y-x$ it follows that $x \mid y$ therefore $y = kx$ and (1) becomes: $x^{(k+1)x}=(kx)^{(k-1)x} \tag2$ and, dividing by $x^{k-1}$ we get
$x^{2x}=k^{(k-1)x} \tag 3$ therefore $x^2=k^{k-1} \tag 4$ and here I've got stuck. Any help is appreciated.
UPDATE
From (4) we have $k \ge 3$ therefore $2 \le k-1$ and $ k \mid x$. Let $x=km$ and (4) becomes $k^2m^2=k^{k-1} \tag5$ and $m^2=k^{k-3} \tag6$
UPDATE 2
$x=t^{t^2-1}, y=t^{t^2+1}$ is solution $\forall t \in \mathbb{N}-\{0\}$
 A: $x^2 = k^{k-1}, k$ must be odd, or k is a perfect square. 
k is odd.
$k = 2n+1\\
x,y = (2n+1)^n, (2n+1)^{n+1}$
$k$ is a pefect square. $k = n^2$
$x^2 = n^{2(2n-1)}\\
x,y = n^{2n-1}, n^{2n+1}$
A: I've written a simple Java program to attempt a bruceforce method:
(I understand this isn't always the best way to solve something like this, but it may help us find patterns)
Here are some solutions I have found:
In the form (x,y): All the solutions between 1 and 1000


*

*(1,1) (not an accepted solution)

*(3,9)

*(8,32)

*(25,125)



Well, well, well...
1/1 = 1
9/3 = 3
32/8 = 4
125/25 = 5


*

*Do I see a pattern? :)

A: From
$x^2=k^{k-1}$,
$k$ must be odd.
Let $k = 2j+1$,
so
$x^2 = (2j+1)^{2j}$
or
$x = (2j+1)^j$.
Then
$y = kx
=(2j+1)(2j+1)^j
= (2j+1)^{j+1}
$.
Check:
$x^{x+y}?y^{y-x}$
$\begin{array}\\
x^{x+y}
&=((2j+1)^j)^{(2j+1)^j+(2j+1)^{j+1}}\\
&=(2j+1)^{j(2j+1)^j(1+(2j+1))}\\
&=(2j+1)^{j(2j+1)^j(2j+2)}\\
&=(2j+1)^{2j(j+1)(2j+1)^j}\\
\end{array}
$
$\begin{array}\\
y^{y-x}
&=((2j+1)^{j+1})^{(2j+1)^{j+1}-(2j+1)^{j}}\\
&=(2j+1)^{(j+1)((2j+1)^{j+1}-(2j+1)^{j})}\\
&=(2j+1)^{(j+1)(2j+1)^{j}((2j+1)-1)}\\
&=(2j+1)^{(j+1)(2j+1)^{j}(2j)}\\
&=(2j+1)^{2j(j+1)(2j+1)^{j}}\\
\end{array}
$
And they match!
A: Here the infinite family of all odd solutions.
$$x^{x+y}=y^{y-x}\iff(xy)^x=(\frac yx)^y\Rightarrow y=kx\\(kx^2)^x=k^{kx}\iff x^{2x}=k^{x(k-1)}\Rightarrow k-1=2n\\x^{2x}=k^{x(k-1)}\Rightarrow x^{2x}=(2n+1)^{2nx}\Rightarrow x=(2n+1)^n\Rightarrow\color{red}{ (x,y)=((2n+1)^n,(2n+1)^{n+1})}$$ This gives the following identity verifying the given equation
$$(2n+1)^{n(2n+1)^n(2n+2)}=(2n+1)^{(n+1)(2n+1)^n(2n)}$$
Something similar way, I guess, must gives the even solutions like $(x,y)=(8,32)$ if there are many of them.
