Find pairs of positive integers $(x, y)$, satisfying
$$
x>y \text {, and }(x-y)^{x y}=x^{y} y^{x} \text {. }
$$
(2013, Taiwan Mathematical Olympiad Training Camp)
Solution 1 Let
$$
(x, y)=d, x=d v, y=d t,(v, t)=1 .
$$

Then $d^{x y-x-y}(v-t)^{x y}=v^{y} t^{x}$
$$
\Rightarrow d^{(x-1)(y-1)-1}(v-t)^{x y}=v^{y} t^{x} \text {. }
$$

If $(x-1)(y-1)-1=-1$
$\Rightarrow x=1$ (discard) or $y=1$
$$
\Rightarrow(x-1)^{x}=x
$$
$\Rightarrow$ no positive integer solution.
If $(x-1)(y-1)-1 \geqslant 0$, then by
$$
(v-t, v)=1,(v-t, t)=1 \text {, }
$$

and $(v-t)^{x y} \mid v^{y} t^{x}$, we know
$$
v-t=1 \text {. }
$$

Thus $d^{d(t+1) t-(2 t+1)}=(t+1)^{t} t^{t+1}$. If $t$ has a prime factor, then for any prime factor $p$ of $t$, let $p^{\alpha}\left\|t, p^{\beta}\right\| d(\alpha, \beta \geqslant 1)$.
Then $\beta[d(t+1) t-2 t-1]=\alpha(t+1)$
$\Rightarrow(\beta d t-\alpha)(t+1)=(2 t+1) \beta$.
Since $(t+1,2 t+1)=1$, hence $(t+1) \mid \beta$.
From equation (1) we know
$$
\begin{array}{l}
\beta \mid \alpha(t+1) \\
\Rightarrow \beta \leqslant \alpha(t+1) \\
\Rightarrow(\beta d t-\alpha)(t+1) \leqslant(2 t+1) \alpha(t+1) \\
\Rightarrow \beta d t \leqslant 2(t+1) \alpha \leqslant 2 \beta \alpha \\
\Rightarrow 2 p^{\alpha} \leqslant d t \leqslant 2 \alpha \Rightarrow p^{\alpha} \leqslant \alpha
\end{array}
$$
$\Rightarrow$ no positive integer solution.
Contradiction.
Therefore, $t=1 \Rightarrow d^{2 d-3}=2 \Rightarrow d=2$.
Thus $(x, y)=(4,2)$.
Solution 2 If $p \mid x, p$ is a prime, then
$$
p|(x-y) \Rightarrow p| y \text {. }
$$

Since $x \geqslant 2$, such a $p$ exists.
Hence $y \geqslant 2$.
Similarly, if $p \mid y$, then $p \mid x$.
Without loss of generality, let $x=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, y=p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}}$.
If there exists a positive integer $i$, such that $\alpha_{i}y \geqslant p_{i}^{\beta_{i}} \geqslant 2^{\beta_{i}}$
$$
\Rightarrow$ no solution.
Thus, for any integer $i(1 \leqslant i \leqslant k)$, we have
$$
\alpha_{i} \geqslant \beta_{i} \Rightarrow y \mid x \text {. }
$$

Let $x=\operatorname{ty}(t \geqslant 2)$. Then
$$
\begin{array}{l}
(t-1)^{t y^{2}} y^{t y^{2}}=t^{y} y^{y+t y} \\
\Rightarrow(t-1)^{t y^{2}} y^{t y^{2}-y-t y}=t^{y} .
\end{array}
$$

If $t \geqslant 3$, then
$$
(t-1)^{t y^{2}}=\left[(t-1)^{y}\right]^{t y}>t^{t y}>t^{y} \text {. }
$$

Since $y^{t y^{2}-y-t y} \geqslant 1$, contradiction.
Thus $t=2 \Rightarrow y^{2 y^{2}-3 y}=2^{y}$
$$
\Rightarrow y^{2 y-3}=2 \Rightarrow y=2, x=4 \text {. }
$$