An arbitrary natural number $N \in \mathbb{N}$ has the divisors $d_1,d_2,...,d_s$ , where $N$ has $s$ divisors in total. If $n_i$ denotes the number of dividers to $d_i, i = 1, ..., s$, it can be proven that
$n_1^{3}+n_2^{3}+n_3^{3}+...+n_s^{3} = (n_1+n_2+n_3+...++n_s)^2$
For example, let $N=6$. $N$ has the divisors $1,2,3,6$ which has the number divisors $1,2,2,4$ respectively. It follows that
$1^3 + 2^3 + 2^3 +4^3 = (1+2+2+4)^2$
stating that
$LHS=RHS$.
Using the fundamental theorem of arithmetic, a pattern emerges. Let $P_i\in \mathbb{P}$, $k_i \in \mathbb{N}$. It follows that any natural number can be written as
$N = P_1^{k_1}\cdot P_2^{k_2}\cdot P_3^{k_3}\cdot ... \cdot P_s^{ks}$.
Using this property, the pattern noted suggests that the greatest divisor's number of divisors can be formulated as
$N = P_1^{k_1}\cdot P_2^{k_2}\cdot P_3^{k_3}\cdot ... \cdot P_s^{ks} \iff n_s=(k_1 +1)(k_2 +1)(k_3 +1)...(k_s +1)$.
This is my progress so far. Any help towards solving this problem would be greatly appreciated.