7. find all natural numbers $n$ such that

$$
\sum_{\substack{d \mid n \\ 1 \leq d<n}} d^{2}=5(n+1)
$$
Solution (Viviane): We consider $n$ even and $n$ odd individually.

Case 1: $n$ is even:

$n$ cannot be 2. If $n>2$, $n$ has the divisors 1,2 and $n / 2$. It follows that $1+4+(n / 2)^{2} \leq$ $5(n+1)$. Transforming gives $n \leq 20$. Testing the cases gives the solution $n=16$.

Case 2: $n$ is odd:

One is not a solution. The right-hand side is even, so the left-hand side is even. All divisors of $n$ are odd. It follows that $n$ has an odd number of divisors (with itself), so $n$ is a square number. Powers of odd prime numbers are not solutions, because then the left-hand side is $\equiv 1 \operatorname{modulo} p$ and the right-hand side is $\equiv 5 \bmod p$. This only works if $0 \equiv 4 \bmod p$, i.e. $p=2$, i.e. $n$ even.

We can therefore write $n=(p s)^{2}$, where $p$ is the smallest prime divisor of $n$ and $s$ has at least one prime divisor $\neq p$. It follows that $n$ has the three different divisors $1, p$ and $p s^{2}$. In addition, $p$ and $s$ are odd and not one, so both are greater than 2. Therefore, the following must apply

$$
5+9(p s)^{2}=1+4+9(p s)^{2} \leq 1+p^{2}+s^{2}(p s)^{2} \leq 5\left((p s)^{2}+1\right)
$$

which is not possible.

So $n=16$ is the only solution.

Other solutions are, for example, if you first limit the number of divisors towards the top and then make case distinctions, or if you estimate the number of different prime divisors and then make case distinctions.

Marking Scheme: Points for different solution paths are not additive, you get points for the solution path where you got furthest.

- Points for MuLö:
- $n$ even: 1 point for $n \leq 20$, then 1 point for $n=16$.
- $n$ odd: 1 point for $n$ square number, then 1 point for $n$ no prime power, 5 points if case $n$ odd in total.
- Points if the number of divisors is estimated:
- less than 16 divisors: 1 point
- less than 8 divisors: 2 points (not additive to 1 point)
- $n$ no prime power: 1 point
- Points if you estimate prime divisors:
- $n$ no prime power: 1 point
- $n$ two different prime divisors: 3 points
- $n$ more than two different prime divisors: 2 points