663. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square.
$\triangle$ For a natural $n$, the inequality holds:

$$
(n+1)^{2} \leqslant n^{2}+3 n<(n+2)^{2}
$$

(check it!). But since the numbers $(n+1)^{2}$ and $(n+2)^{2}$ are the two closest perfect squares to each other, the number $n^{2}+3 n$ will be a perfect square only when the left inequality turns into an equality:

$$
n^{2}+3 n=(n+1)^{2}, \quad n^{2}+3 n=n^{2}+2 n+1, \quad n=1
$$

Answer: 1.

The method of solution we used in problem 663 is called the "pinching" method.