Determine all integers $n \geq 2$, satisfying
$$n=a^2+b^2,$$
where $a$ is the smallest divisor of $n$ different from $1$ and $b$ is an arbitrary divisor of $n$.
[i]Proposed by Walther Janous[/i]
1. We start with the equation \( n = a^2 + b^2 \), where \( a \) is the smallest divisor of \( n \) different from 1, and \( b \) is an arbitrary divisor of \( n \).

2. Since \( a \) is the smallest divisor of \( n \) greater than 1, \( a \) must be a prime number. Let us denote \( a \) as the smallest prime divisor of \( n \).

3. Given that \( b \) is an arbitrary divisor of \( n \), we have \( b \mid n \). This implies that \( b \mid a^2 + b^2 \). Therefore, \( b \mid a^2 \).

4. Since \( b \mid a^2 \) and \( a \) is prime, \( b \) can only be \( 1 \), \( a \), or \( a^2 \).

5. If \( b = 1 \), then \( n = a^2 + 1^2 = a^2 + 1 \). However, \( a \) is the smallest divisor of \( n \) greater than 1, and \( a \) must be a prime number. This implies that \( n \) cannot be expressed as \( a^2 + 1 \) because \( a \) would not be the smallest prime divisor in this case. Therefore, \( b = 1 \) is not a valid solution.

6. If \( b = a \), then \( n = a^2 + a^2 = 2a^2 \). This implies that \( n \) is even, and since \( a \) is the smallest prime divisor, \( a \) must be 2. Substituting \( a = 2 \) into the equation, we get:
   \[
   n = 2 \cdot 2^2 = 8
   \]
   Therefore, \( n = 8 \) is a solution.

7. If \( b = a^2 \), then \( n = a^2 + (a^2)^2 = a^2 + a^4 = a^2(1 + a^2) \). Since \( a \) is prime, if \( a > 2 \), then \( 1 + a^2 \) must be even, making \( n \) even. This would imply that \( n \) is divisible by 2, contradicting the minimality of \( a \). Therefore, \( a \) must be 2. Substituting \( a = 2 \) into the equation, we get:
   \[
   n = 2^2(1 + 2^2) = 4 \cdot 5 = 20
   \]
   Therefore, \( n = 20 \) is another solution.

8. We have found that the only solutions for \( n \) are \( n = 8 \) and \( n = 20 \).

The final answer is \( \boxed{8 \text{ and } 20} \).