Solve for natural numbers: $$x^{y}=y^{x-y}$$
What I have tried:
\begin{equation} x>y \geqslant 2 \end{equation}
\begin{equation} \left(\frac{x}{y}\right)^{y}=y^{x-2 y} \end{equation}
\begin{equation} \frac{x}{y}=k, k \in N \end{equation}
\begin{equation} x=k y \end{equation}
\begin{equation} (k y)^{y}=y^{k y-y} \Leftrightarrow k^{y} y^{y}=y^{k y-y} \Leftrightarrow k^{y}=y^{(k-2) y} \Leftrightarrow k=y^{k-2} . \end{equation}
\begin{equation} k=y^{k-2} \end{equation} \begin{equation} k \geqslant 2^{k-2} \end{equation}
\begin{equation} k<2^{k-2} \end{equation}
\begin{equation} k \geqslant 5 \end{equation}
How do I prove this statement?
The solutions are $(x, y)= (1, 1), (9, 3), (8, 2) $.
$x^y = y^{x-y}$ with $x, y \ge 1$.
If $y=1$ then $x=1$ which is a solution.
Assume $y \ge 2$.
If $x = y$ then $y^y = 1$, so $y = 1$.
If $x \lt y$ then $x^y =1/y^{y-x}< 1$, so no solutions.
Therefore $x > y \ge 2$.
$x =y^{x/y-1}$ so $x/y-1 > 1$ or $x > 2y$.
$(xy)^y = y^x$.
Let $(x, y) = d, x = ad, y=db, (a, b) = 1, a > 2b$.
$(d^2ab)^{db} = (db)^{da},\\ (d^2ab)^{b} = (db)^{a},\\ d^{2b}(ab)^b = d^ab^a,\\ d^{2b}a^b = d^ab^{a-b},\\ a^b = d^{a-2b}b^{a-b}$.
Since $(a, b) = 1$ and $a > 2b$, $b = 1$ (otherwise $p | b\implies p|a$).
Therefore $a = d^{a-2}$ or $d = a^{1/(a-2)}$. Since $a > 2, a \ge 3$.
Let $a = 3+c, c \ge 0$. $d = (3+c)^{1/(1+c)}$.
If $c = 0, d=3, a=3, x=9, y=3$. $9^3 = 3^6$ which works.
If $c = 1$ then $d = 4^{1/2} = 2, a=4, x=8, y=2, 8^2 = 2^6$ which works.
If $c =2$ then $d = 5^{1/3}$ which is not an integer.
If $c \ge 2$ then $1 < (3+c)^{1/(1+c)} < 2$ so there are no solutions.
$3+c \lt 2^{1+c}$ true for $c = 2$. If $2^{1+c} \gt 3+c$ then $2^{1+c+1} =2\cdot 2^{1+c} \gt 2\cdot (3+c) =3+c+3+2c \gt 3+c+1 $.