Problem 1684

1. Consider the sequence and the fractional part:
Let $a$ be a real number. The sequence $n_1 < n_2 < n_3 < \ldots$ consists of all positive integers $n$ such that $\{na\} < \frac{1}{10}$. Here, $\{x\}$ denotes the fractional part of $x$.

2. **Reduction to interval $(0, 1)$:**
We can assume $a \in (0, 1)$ without loss of generality because adding an integer to $a$ does not change the fractional parts $\{na\}$.

3. Interpretation on a unit circle:
Imagine the numbers $\{na\}$ as points on a circle of unit length. We lay off arcs of length $a$ successively from point 0. We are interested in the arrangement of points on a given arc of length $\frac{1}{10}$.

4. Define shifts:
- A shift to the right occurs if $\{n_{k+1}a\} > \{n_k a\}$.
- A shift to the left occurs if $\{n_{k+1}a\} < \{n_k a\}$.
- The magnitude of the shift is $|\{n_{k+1}a\} - \{n_k a\}|$.
- The number of steps of the shift is $n_{k+1} - n_k$.

5. Consecutive shifts in the same direction:
Two consecutive shifts in the same direction are always equal. For example, if $\{n_k a\} < \{n_{k+1} a\} < \{n_{k+2} a\}$, then $n_k < m = n_k + n_{k+2} - n_{k+1} < n_{k+2}$ and $\{m a\} = \{n_k a\} + \{n_{k+2} a\} - \{n_{k+1} a\} < \frac{1}{10}$, implying $m = n_{k+1}$ and $n_{k+2} - n_{k+1} = n_{k+1} - n_k$.

6. Define smallest shifts:
- Let $m_1$ be the smallest number of steps for a right shift $\Delta_1$.
- Let $m_2$ be the smallest number of steps for a left shift $\Delta_2$.
- Assume $m_1 < m_2$.

7. Lemma 1:
If $\Delta_1 + \Delta_2 \geq \frac{1}{10}$ and for some $n$, $\{n a\}$ lies in the interval $[ \frac{1}{10} - \Delta_1, \Delta_2 )$, then there exists a shift of $|\Delta_2 - \Delta_1|$ with number of steps $m_1 + m_2$.

8. Lemma 2:
There are no shifts with a number of steps no more than $m_1 + m_2$, except for the three (or possibly two) mentioned above.

Proof of Lemma 2:
- Suppose there exists a shift of $\Delta$ with number of steps $m$.
- If it is a shift to the right, then $\Delta < \Delta_1$. Find $n_k$ such that $\{n_k a\} < \frac{1}{10} - \Delta_1$. Then $\{(n_k + m) a\} = \{n_k a\} + \Delta < \{n_k a\} + \Delta_1 = \{(n_k + m_1) a\}$. Since $m$ cannot be less than $m_1$, the transition from $\{(n_k + m_1) a\}$ to $\{(n_k + m) a\}$ contains a shift to the left and less than $m_2$ steps, which is impossible.
- If the shift by $\Delta$ is a shift to the left, then $\Delta < \Delta_2$ and $m > m_2$. Taking $n_k$ such that $\{n_k a\} > \Delta_2$, we obtain that the transition from $\{(n_k + m_2) a\}$ to $\{(n_k + m) a\}$ contains a shift to the right and less than $m_1$ steps, which is again a contradiction.

9. Proof of Lemma 1:
- Take a natural $n$ for which $\{n a\}$ lies in the half-interval $[ \frac{1}{10} - \Delta_1, \Delta_2 )$.
- Then $\{(n + m_1 + m_2) a\} = \{n a\} + \Delta_1 - \Delta_2 \in (0, \frac{1}{10})$.
- At no natural $k < m_1 + m_2$ does the fractional part $\{(n + k) a\}$ lie in the interval $(0, \frac{1}{10})$—at $k = m_1$ and $k = m_2$ by the assumption for the number $\{n a\}$, and at the other $k$ by Lemma 2, which has already been proved.

10. Conclusion:
The statement of the problem follows from Lemmas 1 and 2. If $\Delta_1 + \Delta_2 \geq \frac{1}{10}$ and for some $n$ the fractional part $\{n a\}$ lies in the interval $[ \frac{1}{10} - \Delta_1, \Delta_2 )$, then:
- From points with fractional parts smaller than $\frac{1}{10} - \Delta_1$, we have a shift by $\Delta_1$.
- From points with fractional parts greater than $\Delta_2$, we have a shift by $-\Delta_2$.
- The other points are shifted by $|\Delta_2 - \Delta_1|$ (both shifts with fewer steps are impossible).

If no such $n$ exists, there are no points of the last type.

We have proved that if the differences $n_{k+1} - n_k$ take three different values, then one of these values is equal to the sum of the other two.

$\blacksquare$