Does $\lfloor \tan (n)\rfloor =n$ have infinite roots? Background
This post can be understood as the gap between $\tan n$ and $n$ has no upper limit.
Does the sequence $n+\tan(n), n \in\mathbb{N}$ have a lower bound?
I want to know how close $\tan n$ can be to $n$.
I try to find some solution with difference less than 1.
$$|\tan n -n|<1$$
or similar
$$\lfloor \tan (n)\rfloor =n$$
$\lfloor x\rfloor$ means floor function.
I did a quick search for $n<10^7$ and found no solution other than $n = 0,1$
Question
Is there an infinite number of integers n that make the equation true, and how big will the next solution be?


I want to know how close $\tan n$ can be to $n$.

Well, it's exactly equal if $n = 0$, but that's a boring answer.
Otherwise, here are the smallest values of $|\tan n - n|$ that I've found for $|n| \le 10^7$, using simple brute force:

*

*$n = \pm 1$, error ≈ 0.557408.  As noted, $n = +1$ is the only known nonzero solution to $\lfloor \tan n \rfloor = n$.

*$n = \pm 4$, error ≈ 2.842179

*$n = \pm 3$, error ≈ 3.142547

*$n = \pm 2$, error ≈ 4.185040

*$n = \pm 7$, error ≈ 6.128552

*$n = \pm 6$, error ≈ 6.291006

*$n = \pm 14$, error ≈ 6.755393

*$n = \pm 5$, error ≈ 8.380515

*$n = \pm 10$, error ≈ 9.351639

