How do I know that $n \neq k^3$ if $n, k$ are natural numbers and $n$ has exactly 999 divisors? How do I know that $n \neq k^3$ if $n, k$ are natural numbers and $n$ has exactly 999 divisors?
I came across this fact, but it doesn't look intuitive at all. Is there a way to prove this? Presumably, I would need to use some basic number theory, but I'm not really sure where to start.
 A: Suppose, $n$ has the prime factorization $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$. Number of divisors of $n$ can be given by $(\alpha_1+1)(\alpha_2+1)\cdots (\alpha_k+1)$, which is $999$ here. $n$ to be a perfect cube, we need all $\alpha_i=3\cdot x_i$, for some $x_i\in\mathbb{N}$. That means $$(\alpha_1+1)(\alpha_2+1)\cdots (\alpha_k+1)=(3x_1+1)(3x_2+1)\cdots (3x_k+1)$$
But, product of two numbers of the form $(3t+1)$ is also $(3t+1)$, and using induction we have the above number to be of the form $3t+1$. But, $999\equiv 0\pmod{3}$. 
Hence, it is not possible to exist $\alpha_i$ s for which the equation $$(\alpha_1+1)(\alpha_2+1)\cdots (\alpha_k+1)=999$$ holds. 
A: Hint: Prove that if $n=k^3$ for some integer $k$, then the number of divisors of $n$ is one more than a multiple of $3$. (E.g., $2^3$ has $4$ divisors, $6^3$ has $16$ divisors, $100^3$ has $49$ divisors, ...) Since $999$ is not one more than a multiple of $3$, if $n$ has $999$ divisors then it cannot equal $k^3$ for any integer $k$.
To prove this, consider the prime decomposition of $k$. We can write
$$k=p_1^{\alpha_1}\cdots p_t^{\alpha_t}$$
for some distinct primes $p_1,\dots,p_t$. Then a factor of $k^3$ looks like
$$p_1^{\beta_1}\cdots p_t^{\beta_t}$$
where each $\beta_i$ satisfies $0 \leq \beta_i\leq 3\alpha_t$. How many such factors are possible?
