Problem 2403

Consider $n = p_1 \cdots p_{k-1} \cdot p_k^2 \cdot p_{k+1} \cdots p_s$, with $p_1 < p_2 < \dots < p_s$ prime numbers, $1 \le k \le s$, and $p = p_k \equiv 1 \pmod 3$. Moreover, $p$ is odd and satisfies $p^2 + p + 1 \equiv 0 \pmod 3$.

The number of divisors of $n$, $\tau(n)$, is
$$\tau(n) = \underbrace{(1+1) \cdot (1+1) \cdots (1+1)}_{\text{applied } s-1 \text{ times}} \cdot (2+1) = 3 \cdot 2^{s-1},$$
and the sum of the divisors, $\sigma(n)$, is:
$$\begin{aligned} \sigma(n) &= \frac{p_1^2-1}{p_1-1} \cdots \frac{p_{k-1}^2-1}{p_{k-1}-1} \cdot \frac{p_k^3-1}{p_k-1} \cdots \frac{p_{k+1}^2-1}{p_{k+1}-1} \cdots \frac{p_s^2-1}{p_s-1} \\ &= (p_1+1) \cdots (p_{k-1}+1) \cdot (p_k^2+p_k+1) \cdot (p_{k+1}+1) \cdots (p_s+1). \end{aligned}$$

Case I. If $n$ is even, then $p_1 = 2$, and $p_2 < p_3 < \dots < p_s$ are odd primes. Hence, $2\sigma(n) = 2 \cdot (2+1) \cdot (p_2+1) \cdots (p_{k-1}+1) \cdot (p_k^2+p_k+1) \cdot (p_{k+1}+1) \cdots (p_s+1)$, so $2\sigma(n)$ is divisible by $2 \cdot 3^2 \cdot 2^{s-2} = 3 \cdot \tau(n)$.

Case II. If $n$ is odd, then $p_1 < p_2 < \dots < p_s$ are odd primes, so
$$\sigma(n) = (p_1 + 1) \cdots (p_{k-1} + 1) \cdot (p_k^2 + p_k + 1) \cdot (p_{k+1} + 1) \cdots (p_s + 1)$$
is divisible by $3 \cdot 2^{s-1} = \tau(n)$.