Problem 1765

Let $n = 2^a \cdot 3^b p_1^{\alpha_1} \dots p_s^{\alpha_s}$ where $p_i \neq 2, 3$ for $i = 1, 2, \dots, s$ are distinct prime numbers. If $t$ is a divisor of $n$ of the form $6k \pm 1$, then $t$ is a divisor of $p_1^{\alpha_1} \dots p_s^{\alpha_s}$ (in other words, $t$ is not divisible by $2$ or by $3$). Also, all divisors of $p_1^{\alpha_1} \dots p_s^{\alpha_s}$ are of the form $6k \pm 1$. If $g(n)$ and $h(n)$ are of different parity then $g(n) + h(n)$ is odd. Therefore, the number $p_1^{\alpha_1} \dots p_s^{\alpha_s}$ has an odd number of divisors and since the number of divisors equals $(\alpha_1 + 1) \dots (\alpha_s + 1)$ we conclude that $p_1^{\alpha_1} \dots p_s^{\alpha_s}$ is a perfect square. Hence, $n = 2^a \cdot 3^b m^2$.