Simultaneous rational approximations of multiples of the golden ratio

Question

My question concerns potential simultaneous rational approximations of irrational numbers.

Let $\alpha = \frac{1 + \sqrt{5}}{2}$ be the golden ratio, and $k \in \mathbb{N}$ a positive integer. In what follows, for a real number $x$, $|x|$ is the distance from $x$ to $\mathbb{Z}$.

I'm going to try to find good rational approximations of both $\alpha$ and $k\alpha$ with the same denominator. My precise question is the following.

Fix a positive $\epsilon >0$. I was wondering if it were possible to characterise those integers $k$ as above such that there exists a sequence $(n_m)_{m \in \mathbb{N}}$ such that the two followings things hold

1. $|n_m \alpha| \rightarrow 0$
2. $|n_m (k\alpha)| \leq \frac{\epsilon}{n_m}$

A basic remark is that for such a sequence to exist it is necessary that $\epsilon$ be less than $\frac{C}{k}$ for $C$ the constant such that for all $n$ we have $|n \alpha| \geq \frac{C}{n}$.

More precise questions are:

Question 1. Could it be true that for a given $\epsilon$, it is possible to find such a sequence for any $k$ big enough?

Question 2. If the answer to the previous question is no. Given $\epsilon$, can anyone say anything about the density of the set of $k$s such that the property above holds? In particular, does this density tend to zero when $\epsilon$ tend to $0$?

Answer 1

Note that since you want $|n_m \alpha| \to 0$ this implies that for almost all $m$ you have $|n_m \alpha| < \frac{1}{2k}$, and in that case you have $ |n_m k \alpha | = k |n_m \alpha |$. You can have $|n_m \alpha | \leq \frac1{n_m \sqrt5}$, so $\epsilon = \frac{k}{\sqrt 5}$, and this is optimal, because for any $C > \sqrt 5$ you have only a finite number of solutions to $|n_m \alpha| \leq \frac{1}{C n_m}$.

This means that counterintuitively, increasing $k$ actually increases $\epsilon$, not decreases it, which isn't what happens if you don't insist on also approximating $\alpha$, where you can have $\epsilon = \frac1{k\sqrt5}$.