TITLE: How do I prove that there are infinitely many natural numbers $n$ such that $\lfloor\sqrt{3}\cdot\tau(n)\rfloor$ divides $n$?
QUESTION [5 upvotes]: Let $\tau(n)$ denote the number of positive divisors of a natural number $n$. Prove that there are infinitely many natural numbers $n$ such that $\lfloor\sqrt{3}\cdot \tau(n)\rfloor$ divides $n$.
$∗$ A slightly strengthened version: prove that this remains true by replacing $\sqrt{3}$ with an arbitrary irrational number $0<\alpha<2$.
Attempt:

Let $\tau(n)=12$. Then $\lfloor\sqrt{3}\cdot\tau(n)\rfloor=\lfloor\sqrt{432}\rfloor=20$. Consider numbers of the form $n=2^{2}5p$, where $p$ is prime, not equal to $2$ or $5$. Then $\tau(n)=12$, and all such
numbers fit.

Is this the correct solution? (could it be simpler?) And can you please give hints on how to approach the strengthened version?

REPLY [2 votes]: It is known that, for any positive irrational number $\beta > 0$, there exist infinitely many $n \in \mathbb{Z}_{\geq 1}$ such that $\lfloor n\beta \rfloor = p$ is a prime. It is even possible to bound the least prime number occurs in such sequence $\{\lfloor n\beta \rfloor | n \geq 1\}$ (which is called Beatty sequence) - see here for details.
Although the above fact is nontrivial, one can deduce the strengthened version of the problem you mentioned from this. The argument is similar to yours but we use simpler $n$.
For given $\alpha$, let $\beta = 2\alpha$ and one can find $m \geq 2$ with $\lfloor m\beta\rfloor = \lfloor 2m\alpha \rfloor = p$ is prime.
Now, for any prime $q\neq p$, $n = p^{m-1}q$ satisfies $\tau(n) = 2m$ and so $\lfloor \tau(n) \alpha\rfloor = \lfloor m\alpha\rfloor = p $ divides $n = p^{m-1}q$.