http://www.kalva.demon.co.uk/short/soln/sh00n2.html

Peter, your link does not work However i try with the problem.. Let $ n = \prod_i{p_i^{\alpha_i}}$ be the canonical factorization of $ n \in \mathbb{N}$, and $ p_n$ is the $ n$-th prime and $ \alpha_i = \upsilon_{p_n}(n) \ge 0$. The equation becomes $ \prod_i{(\alpha_i + 1)^3} = 4n \implies \alpha_1 \ge 1$. Now there exist some integers $ \beta_1,\beta_2,...$ s.t. $ \alpha_1 = 3\beta_1 - 2, \beta_1 > 0, \alpha_i = 3\beta_i \forall i > 1$. So we obtain $ (3\beta_1 - 1)\prod_{i > 1}{(3\beta_i + 1)} = \prod_i{p_i^{\beta_i}} \implies \beta_2 = 0$. Since $ p^x = p(p^{x - 1}) > pe^{x - 1} \ge p(x - 1) + p = px > 3x + 1, \forall p \ge 5, x \ge 1$ we need to have $ 2^{\beta_1} \le 3\beta - 1 \implies 1 \le \beta_1 \le 3$. If $ \beta_1 = 2 \pm 1 \implies \prod_{i > 2}{(3\beta_i + 1)} \le \prod_{i > 2}{p_i^{\beta_i}}$, with equality iff $ \beta_i = 0, \forall i > 1$. If $ \beta_1 = 2 \implies \beta_3 > 0 \implies (\frac {3\beta_3 + 1}{4 \cdot 5^{\beta_3 - 1}})\prod_{i \ge 4}{(3\beta_i + 1)} \le \prod_{i \ge 4}{p_i^{\beta_i}}$, with equality iff $ \beta_3 = 1, \beta_i = 0, \forall i > 3$. So we can conclude that $ n \in \{2,2^7,2^4 \cdot 5^3\}$. Cheers

After some playing around I think we can show that n can have no more than 2 distinct prime divisors, and this reduces it to studying all numbers of the form $n=2^xp^y$ where $ p $ is a prime not equal to $2$.

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