Divisors...

[SANZEED]

Find all positive integer $m$ such that $d(2m^{3})=2m$ where $d(x)$ means the number of divisors of $x$.

[Fm Jakaria]

Write
$m = 2^{s} \prod^{i = k}_{i=1} {p_{i}}^{e_{i}}$ be the prime factorization of $m; s, k$ are nonnegative; all $p_i$’s divide m.
From the given relation:
$(3s + 2) \prod (3e_i+1) = 2^{s+1} \prod {p_i}^{e_i} = 2m….(1)$
It tells that $3$ does not divide $m$.

Now simply note that
1. $2^x \geqslant 3x – 1; \forall x \in \mathbb{N}, x \neq 2$; equality iff $x = 1 or 3$.
2. $p^j > 3j + 1; \forall$ prime $p > 3$ and $\forall j \in \mathbb{N}$.

Note we can apply $1$ with $x = s + 1$; remembering $2$. Now two cases:
1. $s = 0$ or $2$; and $k = 0 \rightarrow m = 1$ or $4$.
2. $s = 1$ and $k > 0$. Then $p_1 = 5$ and
$\prod (3{e_i}+1) = 4* 5^{{p_1} – 1} \prod_{i = 2}^{i = k} {p_i}^{e_i}.$
Use $4. 5^{t-1} \geqslant 3t + 1; \forall t \in \mathbb{N}$ equality iff $t = 1$. Combine with $2$. We get
$i = 1$ and $e_1 = 1 \rightarrow m = 10$.

So $m = 1, 4, 10$ are the solutions.