# 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.

• I think you mean: how do I know such a number n can exist, as not all n that have the first property will have the second. – user645636 Apr 1 '19 at 17:38
• @RoddyMacPhee: to be honest, I think Masie's formulation is clearer than yours. – TonyK Apr 1 '19 at 17:38
• Then you don't know nath @TonyK – user645636 Apr 1 '19 at 18:11
• @RoddyMacPhee: yo mama don't know nath! – TonyK Apr 1 '19 at 18:31
• my mothers dead. anyways I give up. – user645636 Apr 1 '19 at 18:32

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.
• Yes, it's a standard formula in ENT. It's known as $\tau$(tau) function, see this post. – tarit goswami Apr 1 '19 at 18:40
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?