<s>[INST] 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 $\forall t \in \mathbb{N}-\{0\}$
 [/INST] $x^2 = k^{k-1}, k$ 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. $k = n^2$
$x^2 = n^{2(2n-1)}\\
x,y = n^{2n-1}, n^{2n+1}$
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