IMC 2014 · Day 1 · P4

Suppose that $e_1$ is odd, contrary to the statement.

We know that $s(n) = \prod\limits_{i=1}^{k} (1 + p_i + p_i^2 + \dots + p_i^{e_i}) = 2n = 2 p_1^{e_1} \cdots p_k^{e_k}$. Since $e_1$ is an odd number, $p_1 + 1$ divides the first factor $1 + p_1 + p_1^2 + \dots + p_1^{e_1}$, so $p_1 + 1$ divides $2n$. Due to $p_1 + 1 > 2$, at least one of the primes $p_1, \dots, p_k$ divides $p_1 + 1$. The primes $p_3, \dots, p_k$ are greater than $p_1 + 1$ and $p_1$ cannot divide $p_1 + 1$, so $p_2$ must divide $p_1 + 1$. Since $p_1 + 1 < 2 p_2$, this possible only if $p_2 = p_1 + 1$, therefore $p_1 = 2$ and $p_2 = 3$. Hence, $6 | n$.

Now $n$, $\frac{n}{2}$, $\frac{n}{3}$, $\frac{n}{6}$ and $1$ are distinct divisors of $n$, so $$s(n) \ge n + \frac{n}{2} + \frac{n}{3} + \frac{n}{6} + 1 = 2n + 1 > 2n,$$ contradiction.

Remark. It is well-known that all even perfect numbers have the form $n = 2^{p-1} (2^p - 1)$ such that $p$ and $2^p - 1$ are primes. So if $e_1$ is odd then $k = 2$, $p_1 = 2$, $p_2 = 2^p - 1$, $e_1 = p - 1$, and $e_2 = 1$. If $n > 6$ then $p > 2$ so $p$ is odd and $e_1 = p - 1$ should be even.