Equality case in Dirichlet's approximation theorem

Question

In its strong form, Dirichlet's approximation theorem (for dimension one) states that for $\alpha\in{\mathbb R}$, and an integer $n\geq 1$, we have $$ \min_{1\leq k\leq n}\| k \alpha \| \leq \frac{1}{n+1} $$ where $\|x\|$ denotes distance from $x$ to the closest integer.
This strong form is shown here on MSE for example.

My question: how to prove or disprove that equality is reached (in other words $\min_{1\leq k\leq n}\|k \alpha\|$ is equal to $\frac{1}{n+1}$) iff $\alpha$ is a reduced fraction with denominator $n+1$ ?

One of the directions is easy : if $\alpha=\frac{j}{n+1}$ with $j$ coprime to $n+1$, then there is a $k\in[1,\ldots, n]$ such that $jk\equiv 1 (\mod n)$, and we then have $\| k\alpha \| = \frac{1}{n+1}$ ; furthermore, all the other $\lbrace k'\alpha \rbrace$ for $k'\neq k$ are (not necessarily reduced) positive fractions with denominator $n+1$, so that $\|k'\alpha\| \geq \frac{1}{n+1}$.

Update 01/02/2023 In the other direction, suppose that $\min_{1\leq k\leq n}\| k \alpha \|=\frac{1}{n+1}$ and that this min is attained at some $k_0\in [1,\ldots, n]$. Then $\|k_0 \alpha \|=\frac{1}{n+1}$ means that there is an integer $q$ such that $k_0\alpha = q \pm \frac{1}{n+1}$, or $\alpha=\frac{q(n+1)\pm 1}{k_0(n+1)}$. The goal is then to show that either $k_0$ divides the numerator, or else $\|k\alpha\| \lt \frac{1}{n+1}$ for some $k$. We have thus reduced the problem to a purely arithmetic question, and I'm therefore adding the [modular-arithmetic] tag.

Answer 1

Here is a proof for the "only if" direction.

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Suppose for $\alpha\in{\mathbb R}$, and an integer $n\geq 1$, we have $\min_{1\leq k\leq n}|| k \alpha || = \frac{1}{n+1}$.

Let ${x}$ be the fractional part of $x$, i.e., ${x}=x-\lfloor x\rfloor$. Consider $\{\alpha\},\{2\alpha\},\cdots,\{n\alpha\}$. Sort them into a (weakly) increasing sequence $a_1,a_2,\cdots,a_n$. In other words, $a_i\le a_{i+1}$. $\ a_i=\{k_i\alpha\}$, where $k_1,k_2, \cdots, k_n$ is a permutation of $1,2,\cdots, n$.

Since $a_{i+1}-a_i=(k_{i+1}\alpha-\lfloor k_{i+1}\alpha\rfloor)-(k_i\alpha-\lfloor k_i\alpha\rfloor)=(k_{i+1}-k_i)\alpha-b_i$, where $b_i=\lfloor k_{i+1}\alpha\rfloor-\lfloor k_i\alpha\rfloor$ is an integer, we have $$\|a_{i+1}-a_i\|=\|(k_{i+1}-k_i)\alpha\|$$ Since $1\le |k_{i+1}-k_i|\le n$, $\|(k_{i+1}-k_i)\alpha\|=\|\,|k_{i+1}-k_i|\alpha\,\|$$ \ge\frac1{n+1}$. Since $a_{i+1}\ge a_i$, we must have $a_{i+1}-a_i\ge\frac1{n+1}$, for all $1\le i\le n-1$.

Also note that $a_1\ge\frac1{n+1}$ and $1-a_n\ge\frac1{n+1}$. We have $$\begin{aligned} 1&=a_1+(a_2-a_1)+(a_3-a_2)+\cdots+(a_n-a_{n-1})+(1-a_n)\\ &\ge (n+1)\frac1{n+1}\\ &=1 \end{aligned}$$ Hence, $a_1=a_2-a_1=a_3-a_2=\cdots=a_n-a_{n-1}=1-a_n=\frac1{n+1}.$

So $a_i=\frac i{n+1}$ for all $i$. In particular, $\{\alpha\}=\frac m{n+1}$ for some integer $m$. So, $\alpha=\frac {m'}{n+1}$, where $m'=m+\lfloor \alpha\rfloor(n+1)$ is an integer.

$m'$ and $n+1$ must be coprime. Otherwise, $\|((n+1)/d)\alpha\|=0<\frac1{n+1}$, where $d=\gcd(m', n+1)>1$, which contradicts with the given condition since $(n+1)/d$ is an integer between $1$ and $n$.