European Mathematical Cup

Alternative way to define $f(n)$ is
$$f(n) = \sum_{k|n, k \ge 1} d(k).$$
Let $n = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_r^{\alpha_r}$ be the prime factorization of $n$. We have $d(n) = \prod_{i=1}^r (\alpha_i + 1)$.
We prove the function $f$ is multiplicative, in particular, given coprime $n, m$ we have $f(mn) = f(m)f(n)$.
Using $n, m$ are coprime for the second inequality and the fact that function $d$ is multiplicative we get:
$$f(mn) = \sum_{k|m} d(k) = \sum_{k_1|n, k_2|m} d(k_1k_2) = \sum_{k_1|n} d(k_1)d(k_2) = \left(\sum_{k_1|n} d(k_1)\right)\left(\sum_{k_2|m} d(k_2)\right) = f(n)f(m).$$
If $r=1$ we have $n = p_1^{\alpha_1}$. We note that divisors of $n$ are $1, p_1, p_1^2, \dots, p_1^{\alpha_1}$, so
$$f(n) = \sum_{i=0}^{\alpha_1} (i+1) = \frac{(\alpha_1+1)(\alpha_1+2)}{2}.$$
Combining this with the multiplicativity result for $f$ we deduce $f(n) = \prod_{i=1}^n \frac{(\alpha_i+1)(\alpha_i+2)}{2}$.
We now prove that for primes $p \ge 5$ and $p=3$ provided $a \ge 3$ we have $f(p^a) = \frac{(a+1)(a+2)}{2} < \frac{2}{3}p^a$ by induction on $a$. As a basis $3 < \frac{2p}{3}$ for $p \ge 5$ and $6 < \frac{2}{3} \cdot 3^3$. For the step it is enough to notice that $\frac{a+3}{a+1} \le 2 < p$ in both cases.
Similarly we can prove for $p=2$ that $f(p^a) < p^a$ provided $a \ge 4$. By explicitly checking the remaining cases $p=2$ and $a=1,2,3$ and $p=3, a=1,2$ we conclude $f(p^a) \le \frac{2}{3}p^a$ for all $p, a$ and $f(p^a) \le p^a$ for all $p \ge 3$ and $p=2, a \ge 4$.
Assuming $f(n)=n$ we would have $\prod_{i=1}^k \frac{f(p_i^{\alpha_i})}{p^{\alpha_i}} = 1$ so the above considerations imply that only possible prime divisors are 2,3.
If $k=1$ the only possible solution is $n=3$. If $k=2$ we have $p_1=2, p_2=3$ and $1 \le a_1 \le 2$ and $1 \le a_2 \le 2$ which give 4 cases to check giving the other 2 solutions $n=18,36$.

We hereby present one similar but different solution which does not use a lot of properties of the function $f$.
Firstly, we will prove the following lemma:
Lemma. For any positive integer $n>1$ and prime $p$ we have
$$f(pn) \le 3f(n).$$
The equality holds if and only if $GCD(p,n)=1$.
Proof. For every integer $m$ we have that the set of divisors of the number $pm$ is the union of the following two sets:
* set of divisors of $m$
* set of divisors of $m$ multiplied by $p$.
Also those two mentioned sets are disjoint if and only if $GCD(p,m)=1$ (if we have that $p,m$ are disjoint, then it is obvious that none of the divisors of $pm$ are in both sets; if they are not coprime, then the number $p$ belongs to both sets).
This is why we have $d(pn) \le 2d(m)$ and
$$f(pn) = \sum_{k|pn} d(k) \le \sum_{k|n} d(k) + \sum_{k|n} d(pk) \le f(n) + \sum_{k|n} 2d(k) = 3f(n).$$
In both inequalities equality holds if and only if sets from before are disjoint, i.e. when $GCD(p,n)=1$.

Also, we simply see that $f(2^k) = d(1) + d(2) + ... + d(2^k) = 1 + 2 + 3 + ... + (k+1) = \frac{(k+1)(k+2)}{2}$.
Notice that if for some positive integer $n$ we have $f(n) < n$, then for every $p \ge 3$ we have $f(pn) \le 3f(n) \le pf(n) < pn$. Consequently, if $f(n) < n$, then for every odd $m$ we have $f(mn) < mn$.
Because of this, we will introduce new terms. Number $n$ is nice multiple of $m$ if $m|n$ and $m/n$ is odd number. Analogously, we define nice divisor. Our statement from above is: if for some $n$ we have $f(n) < n$, then neither of its nice multiplies is almost perfect number.
Our strategy will be the following: check the cases of the small numbers and see ratio of numbers $n$ and $f(n)$. When we have that $n > f(n)$, conclude that there are not almost perfect numbers among their nice multiplies. With formula for $f(2^k)$ conclude that for sufficiently big $k$ (when $f(2^k) < 2^k$ this is enough to conclude that there are no more almost perfect numbers.
By induction, it is simple to prove that $f(2^k) < 2^k$ for $k \ge 4$. Thus, there are no almost perfect numbers of the form $2^k \cdot m$, where $k \ge 4$ and $m$ is odd, since they all have $2^k$ as their nice divisor. We only have to check the numbers of the form $2^k \cdot m$, where $k \le 3$ and $m$ is odd.
**First case: $k=0$**
For any odd prime $p$ we have $f(p) = d(1) + d(p) = 3 \le p$. From that we see that $n=3$ is solution. Moreover, we do not have any more solutions: if some odd number has a prime divisor different from 3, since $f(p) < p$ this number can not be almost perfect number; if it is a power of 3 bigger than 3, since $f(9) < 3f(3) = 9$, there are no more solutions as well (9 is nice divisor of every power of 3 bigger that 3).
**Second case: $k=1$**.
For any odd prime we have $f(2p) = 3f(2) = 9$. If $p > 5$ then we have $2p > f(2p)$, so for all almost perfect numbers of the form $2^1m$ number $m$ has to have prime divisors 3 and/or 5.
We directly see that neither 6 or 10 is almost perfect. SO, in this case, almost perfect number has to be a nice divisor of the form $2 \cdot 9, 2 \cdot 15$ or $2 \cdot 25$. For $n=18$ we have another solution, in other two cases we have inequality $f(n) < n$. If we want to seek new solution in this case, since they cannot be nice multiplies of 30 and 50, the only possibility is that almost perfect number has nice divisor $2 \cdot 27$. But we have (equality case in lemma) that $f(2 \cdot 27) < 3f(2 \cdot 9) = 2 \cdot 27$. So, there are no more solutions in this case.
**Third case: $k=2$**
For any odd prime we have $f(4p) = 3f(4) = 18$. If $p > 5$ then we have $4p > f(4p)$, so for all almost perfect numbers of the form $2^1m$ number $m$ has to have prime divisors 3 and/or 5.
We directly see that neither 12 or 20 is almost perfect. So, in this case, almost perfect number has to have a nice divisor of the form $4 \cdot 9, 4 \cdot 15$ or $4 \cdot 25$. For $n=36$ we have another solution, in other two cases we have inequality $f(n) < n$. If we want to seek new solution in this case, since they cannot be nice multiplies of 60 and 100, the only possibility is that almost perfect number has nice divisor $4 \cdot 27$. But we have (equality case in lemma), that $f(4 \cdot 27)<3f(4 \cdot 9)=4 \cdot 27$.
So, there are no more solutions in this case.
Fourts case: $k=3$
For any odd prime we have $f(8p)=3f(8)=30$. Similarly to other cases, we only observe candidates of the form $8\cdot 3^l$. Number $8\cdot$ 3 is not almost perfect, all other candidates have nice divisor $8\cdot$ 9. But, we have $f(72)=60<72$. As we always concluded, we do not have any new solutions.
So all almost perfect numbers are 3,18,36.