Determine all natural numbers $n \geq 2$ for which

$$
n=a^{2}+b^{2}
$$

holds, where $a$ is the smallest divisor of $n$ different from 1 and $b$ is any divisor of $n$.

(Walther Janous)
(Walther Janous) We distinguish three cases for $b$.

1. $b=1$. Then $n=a^{2}+1$. From $a \mid n$, that is $a \mid a^{2}+1$, it follows $a \mid 1$, thus the contradiction $a=1$.
2. $b=a$. Then $n=2 a^{2}$ with $a$ being prime. Since $n$ is even, $a$ must be $2$, leading to $n=8$.
3. $b>a$. From $n=a^{2}+b^{2}$ and $b \mid n$, it follows that $b \mid a^{2}$, hence $b=a^{2}$ (since $a$ is prime). Therefore: $n=a^{2}\left(a^{2}+1\right)$ As the product of two consecutive numbers, $n$ is even, so $a$ must be $2$. This results in $n=20$.