China Girls' Mathematical Olympiad

If $x=1$, then $y=1$. If $y=1$, then $x=1$. If $x=y$, then $x^y=1$, so $x=y=1$.

We will discuss the circumstances when $x>y\ge 2$ below. By assumption $$1<{\left(\frac{x}{y}\right)}^y=y^{x-2y},$$ so $x>2y$, and $y\mid x$.

Assume $x=ky$ then $k\ge 3$, and $$ky=y^{(k-2)y}.$$ Therefore $$k=y^{k-2}.$$ Since $y\ge 2$, so $y^{k-2}\ge 2^{k-2}$. By mathematical induction or the binomial theorem, it is easy to prove that when $k\ge 5$, $2^k>4k$. Hence $k$ can only be 3 or 4. When $x=9$, $y=3$, we have $k=3$. When $y=2$, $x=8$, we have $k=4$. Therefore, all pairs of positive integers to be found are $(1,1)$, $(9,3)$ and $(8,2)$.