China-TST-2023B

Proof 1. Nonexistence. Assume that there exists such a positive integer $n$. Let $x$ be an irrational number. Since the terms of the sequence $\{x\}$, $\{2x\}$, $\dots$ are distinct and densely distributed in the interval $(0,1)$, there exist infinitely many positive integers $d$ satisfying $$\{dx\}=\min \{\{ax\}\mid a=1,2,\dots ,d\}.$$ Arrange these $d$ in an infinite increasing sequence $d_1<d_2<\dots$. We will now show that for all $n=d_i-1$ ($i\ge 2$), we always have $$\min \{\{kx\}\mid k=1,\dots ,n\}=\{d_{i-1}x\}\ge \frac{1}{d_i}.$$ Suppose that this inequality does not hold, then $d_i\cdot \{d_{i-1}x\}\le 1$. Thus, $$\{d_{i-1}{d}_{i}x\}=\{d_i\cdot \lfloor d_{i-1}x\rfloor +d_i\cdot \{d_{i-1}x\}\}=d_i\cdot \{d_{i-1}x\}.$$ On the other hand, $$\{d_i{d}_{i-1}x\}=\{d_{i-1}\lfloor d_{i}x\rfloor +d_{i-1}\{d_{i}x\}\}\le d_{i-1}\{d_{i}x\}.$$ Combining the above two equations, we obtain $d_{i-1}\cdot \{d_{i}x\}\ge d_i\cdot \{d_{i-1}x\}$. However, $d_{i-1}<d_i$ and $\{d_{i}x\}<\{d_{i-1}x\}$, which leads to a contradiction! Therefore, $n=d_i-1$ satisfies the given condition, and there are infinitely many such $n$. $\square$

Proof 2. Nonexistent. If there is no irrational number $x$ satisfying the condition, then there exists $n_0\in {\mathbb{Z}}_{+}$ such that for any $n\ge n_0$, there exists $k\in \{1,\dots ,n\}$ such that $\{kx\}<\frac{1}{n+1}$. Since $x$ is an irrational number, for any $k\in {\mathbb{Z}}_{+}$, $\{kx\}\ne 0$. For $n_0$, there exist $k_0\in \{1,\dots ,n_0\}$ and ${\ell }_0\in \mathbb{Z}$ such that $$0<k_{0}x-{\ell }_0<\frac{1}{n_0+1}.$$ Without loss of generality, assume that $k_0$ and ${\ell }_0$ are coprime. Otherwise, we can replace $k_0$ and ${\ell }_0$ with $\frac{k_0}{\gcd (k_0,{\ell }_0)}$ and $\frac{{\ell }_0}{\gcd (k_0,{\ell }_0)}$ respectively, and the condition still holds. Let $n_1=\lfloor \frac{1}{k_{0}x-{\ell }_0}\rfloor >n_0$. Then there exist $k_1\in \{1,\dots ,n_1\}$ and ${\ell }_1\in \mathbb{Z}$ such that $$0<k_{1}x-{\ell }_1<\frac{1}{n_1+1}<k_{0}x-{\ell }_0.$$

Similarly, we can assume that $k_1$ and ${\ell }_1$ are coprime. Since $k_0$ is coprime with ${\ell }_0$ and $k_1$ is coprime with ${\ell }_1$, we have $\frac{{\ell }_0}{k_0}\ne \frac{{\ell }_1}{k_1}$. Therefore, $$\begin{array}{rl}1\le |k_0{\ell }_1-k_1{\ell }_0|& =|k_1(k_{0}x-{\ell }_0)-k_0(k_{1}x-{\ell }_1)|\\ & <\max \{k_1(k_{0}x-{\ell }_0),k_0(k_{1}x-{\ell }_1)\}(\text{because }{k}_1(k_{0}x-{\ell }_0)>0,k_0(k_{1}x-{\ell }_1)>0)\\ & \le \max \left\{\frac{k_1}{n_1},\frac{k_0}{n_1+1}\right\}\le 1(\text{because }\frac{1}{k_{0}x-{\ell }_0}\ge n_1,\text{ hence }{k}_{0}x-{\ell }_0\le \frac{1}{n_1}.)\end{array}$$ This leads to a contradiction. $\square$

Proof 3. For any irrational number $x$, the fractional part $\{kx\}$ is distinct and densely distributed in the interval $(0,1)$. Hence, there exist infinitely many positive integers $m$ satisfying: $$\{mx\}<\{kx\},\forall k=1,2,\dots ,m-1.$$ For each such $m\ge 2$, let $\beta$ be the smallest value among $\{x\}$, $\{2x\}$, ..., $\{(m-1)x\}$. Thus, for $k=1,2,\dots ,m-1$, we have $\{kx\}\ge \beta$, which implies that there are no integers in the open intervals $(kx-\beta ,kx)$. Consider the points $O:(0,0)$, $A:(m,mx)$, $B:(m,mx-\beta )$, $C:(0,-\beta )$ on the coordinate plane. The parallelogram $OABC$ does not contain any lattice points in its interior, and on its boundary, there are exactly three lattice points: $O$, $D:(m,\lfloor mx\rfloor )$, and $E:(r,rx-\beta )=(r,\lfloor rx\rfloor )$. The lattice triangle $△ODE$ has no lattice points inside or on its boundary, except for the vertices. By Pick's theorem, its area is $\frac{1}{2}$. Therefore, the area of the parallelogram $OABC$ is $\ge 2S_{△ODE}=1$. On the other hand, the area of this parallelogram is $m\times \beta =m\times \{rx\}\ge 1$. Hence, $\beta \ge \frac{1}{m}$, and for $k=1,2,\dots ,m-1$, we have $$\{kx\}\ge \{rx\}\ge \frac{1}{m}.$$ Therefore, $n=m-1$ satisfies the given condition. There are infinitely many such $n$, implying that no irrational number $x$ satisfies the condition. $\square$