3. let $n$ be a positive integer with at least four different positive divisors. Let the four smallest of these divisors be $d_{1}, d_{2}, d_{3}, d_{4}$. Find all such numbers $n$ for which

$$
d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=n
$$

## Solution
It is $d_{1}=1$. Suppose $n$ is odd, then $d_{1}, d_{2}, d_{3}, d_{4}$ would also be odd and therefore $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$ even, contradiction. So $n$ is even and $d_{2}=2$. Suppose $n$ is divisible by 4, then one of the numbers $d_{3}, d_{4}$ is equal to 4 and the other is equal to $a$, where $a$ is any natural number. Now 0 and 1 are the only quadratic remainders mod 4, therefore modulo 4 applies

$$
0 \equiv n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}=1^{2}+2^{2}+4^{2}+a^{2} \equiv 1+a^{2} \equiv 1,2
$$

Contradiction. Therefore $n$ is not divisible by 4 and thus $d_{3}=p$ is an odd prime number. For $d_{4}$ the possibilities remain $d_{4}=2 p$ or $d_{4}=q$ with an odd prime number $q>p$. Assuming the latter is the case, then modulo 4 applies

$$
2 \equiv n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2} \equiv 1^{2}+2^{2}+p^{2}+q^{2} \equiv 1+0+1+1 \equiv 3
$$

Contradiction. Therefore, $d_{4}=2 p$ and $n=1^{2}+2^{2}+p^{2}+(2 p)^{2}=5\left(1+p^{2}\right)$. Now $p$ is a divisor of $n$ and also not related to $1+p^{2}$. Therefore, $p$ must be a divisor of 5 , i.e. equal to 5. The only solution is therefore $n=130$.