8.1. Solve the equation $x^{x+y}=y^{y-x}$ in natural numbers. In your answer, specify the sum $x+y$ for the solution $(x, y)$, where $y$ is the smallest, exceeding 1500.
Answer: 2744.

Solution. Let $y = t x$, where $t \in \mathbb{Q}$ for $x, y \in \mathbb{N}$. Then, substituting this into the equation, we find

$$
x = t^{\frac{t-1}{2}}, \quad y = t^{\frac{t+1}{2}}
$$

We will show that the number $t$ can only be an integer. Suppose the contrary: let $t$ be a rational number different from an integer, i.e., $t = \frac{p}{q}$, where the numbers $p$ and $q$ are coprime, $q \neq 1$. Then $y = \frac{p x}{q}$, and the equation becomes

$$
x^{x + \frac{p x}{q}} = \left(\frac{p x}{q}\right)^{\frac{p x}{q} - x} \Leftrightarrow x^{x q + p x} = \left(\frac{p x}{q}\right)^{p x - q x} \Leftrightarrow x^{x q} = p^{p x - x q} \cdot x^{-q x} \cdot q^{x(q - p)} \Leftrightarrow x^{2 q x} = \left(\frac{p}{q}\right)^{x(p - q)}
$$

In the last equation, the left side is always an integer, while the right side is not an integer. Contradiction.

Thus, $t \in \mathbb{N}$, so the numbers $x$ and $y$ will be natural only in the following two cases:

1) the number $t$ is odd: $t = 2k - 1$, then $x = (2k - 1)^{k - 1}, y = (2k - 1)^{k}, k \in \mathbb{N}$;
2) the number $t$ is a perfect square: $t = k^2$, then $x = k^{k^2 - 1}, y = k^{k^2 + 1}, k \in \mathbb{N}$.

We will summarize the obtained solutions in a table:

| $t$ | 1 | 3 | 4 | 5 | 7 | 9 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 1 | 3 | 8 | 25 | 343 | 6561 | $\ldots$ |
| $y$ | 1 | 9 | 32 | 125 | 2401 | 59049 | $\ldots$ |

The solution $(x, y)$, in which $y$ is the smallest, exceeding 1500, is $x = 343, y = 2401$, in which case $x + y = 2744$.