22. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square.
22. For $n=1$ the number $n^{2}+3 n$ equals 4, i.e., is a perfect square. Let now $n>1$. Obviously, in this case

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

Since a perfect square cannot exist between the squares of two consecutive natural numbers, the problem has a unique solution $n=1$.