Problem 759

## Solution.

The given equation can be written in the form

$$\begin{gathered} 5^{n}=3^{2 n}-2 \cdot 3^{n} \cdot 2^{n}+2^{2 n} \\ 5^{n}=\left(3^{n}-2^{n}\right)^{2} \end{gathered}$$

The right-hand side is a perfect square, so the left-hand side must also be a perfect square.

Therefore, $n$ is an even number, $n=2 k(k \in \mathbb{N})$.

$$\begin{gathered} 5^{k}=3^{2 k}-2^{2 k} \\ 5^{k}=\left(3^{k}-2^{k}\right)\left(3^{k}+2^{k}\right) \end{gathered}$$

The difference between the numbers $3^{k}-2^{k}$ and $3^{k}+2^{k}$ is $2^{k+1}$, so it is not possible for both numbers to be divisible by 5. 2 points Therefore, the only possibility is $3^{k}-2^{k}=1$ and $3^{k}+2^{k}=5^{k}$, hence $k=1$. 2 points

The only solution to the given equation is $n=2 k=2$.