Balkan Mathematical Olympiad Shortlist

Let $n=2^a\cdot 3^b{p}_1^{{\alpha }_1}\dots p_s^{{\alpha }_s}$ where $p_i\ne 2,3$ 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, $n=2^a\cdot 3^b{m}^2$.