## On the lonely runner conjecture III

We consider (LRC) .

Conjecture Suppose $n>1$ runners having distinct constant integer speeds $v_k$ start at a common point (origin) and run laps on a circular track with circumference 1. Then, there is a time when no runner is closer than $\frac{1}{n+1}$ from the origin.

Case of $n=2,3$ runners is discussed here. Here we look at case $n=4$.

Let $\| x \|$ denote the distance of x to the nearest integer.

Case $v_k \mod 5 \neq 0,\ \forall k$. At time $t=\frac{1}{5}$ all runners are at least $\frac{1}{5}$ from the origin. Done.

Case $v_k= 5 r_k, k=1..3, v_4 \mod 5 \neq 0$. There is time $t_1$ such that runners $r_k, k=1..3$ are at least $\frac{1}{4}$ from the origin – see case $n=3$. At time $t_2= \frac{t_1}{5}$ runners $v_1, v_2, v_3$ are at the same positions as runners $r_1, r_2, r_3$ at time $t_1$. If at time $t_2$, $\| v_4 t_2\| \geq \frac{1}{5}$ we are done. Otherwise, since $v_4$ and 5 are co-prime, $\exists m : \| v_4 \frac{m}{5} \|=$ $\frac{2}{5} \Rightarrow$ $\| \frac{2}{5} - v_4 \left( t_2+\frac{m}{5} \right) \| \leq \frac{1}{5}$. Done.

Case $v_k= 5 s_k, k=1,2, v_k \mod 5 \neq 0, k=3,4$.

Definition Sector $S_k= \left( \frac{k}{5}, \frac{k+1}{5} \right), k=0..4$.
There is a time $t_1$ such that runners $s_k, k=1,2$ are at least $\frac{1}{3}$ from the origin – see case $n=2$. Since $v_k, k=3,4$ and 5 are co-prime those runners visit all sectors once at times $\frac{m}{5}, m=0..4$, and runners 1 and 2 will be at the same position. There 5 such times, during 2 times runner 3 will be at sectors $S_0, S_4$ and during 2 times runner 4 will visit the same sectors. Therefore, there is $m:\ \| v_k \left( \frac{t_1+m}{5} \right) \| > \frac{1}{5}$. Done.

Two previous cases follow from this post.
Now difficult case.

Case $v_1= 5 s_1,\ v_k \mod 5 \neq 0, k=2,3,4$.

Rearrange speeds that $v_2+v_3 = 5 s_2$. If there more than one way choose the one with maximum sum. We have 2 sub-cases: $s_1 \geq s_2$, $s_1< s_2$.

1. $s_1 \geq s_2$
$\exists m:\ \| v_2 \frac{m}{5} \|=\| v_3 \frac{m}{5} \| = \frac{2}{5}$. The runner 1 is faster than either runner 2 or 3. Therefore, they meet at distance greater than $\frac{1}{5}$.

Now runner 4.

1. $\| v_4 \frac{m}{5} \| = \frac{2}{5} \Rightarrow$ this runner is slower than either runner 2 or 3 (call this runner $r$) which is at the same position as runner 4. Therefore, it will meet runner 1 later than runner $r$, which meet runner 1 at distance larger than $\frac{1}{5}$. Done.
2. $\| v_4 \frac{m}{5} \| = \frac{1}{5} \Rightarrow$. This runner should move fast to reach runner 1 at distance $\frac{1}{5}$. So, it will cover at least $\frac{3}{5} + l$ of the lap. Then, looking at opposite direction in time in the same time it will cover the same distance, and will be at least $\frac{2}{5}$ from the origin when runner 1 reach distance $\frac{1}{5}$ If it is moving even faster to meet runner 1 at $\frac{1}{5}$ it will miss runner 1 from the opposite side. Namely, it has to cover at least the whole lap, and therefore meet runner 1 at the opposite side before runner 1 reaches $\frac{1}{5}$. Done.
2. $s_1< s_2$ This is the most interesting case.

At times $t=\frac{m}{5s_2}, m=1..5s_2-1$ runners 2 and 3 are the same distance from the origin on the opposite sides from the origin.

Runner 1 exhibit repeated motion at $t$. It makes $5 \gcd ( s_1,s_2)$ cycles, where $\gcd$ stands for greatest common divisor. The number of different positions it attain evenly distributed along the track is $n_1=\frac{s_2}{\gcd ( s_1,s_2)} \geq 2$. If $n_1=2$ half of the time runner 1 is away from the origin. With larger $n_1$ there is bigger fraction, with the limit $\frac{3}{5}$. Let $t_1 \in t$ is the time when runner 1 is distant. At the same time runners 2 and 3 same distance on the opposite sides from the origin. Now at times $t_1+\frac{m}{5}, m=0..4$ runner 1 stays at the same position, runners 2 and 3 are on the opposite position relative to the origin, and runner 4 visiting all sectors once each time. Therefore, there are total 2 moments when runner 4 in $S_0, S_4$, and there are total, possibly different 2 times runner 2 in $S_0, S_4$, on the other hand when runner 2 in sector 0 runner 3 is in sector 4 and vice verse. Therefore, there is $m:\ \| v_k \left( t_1 + \frac{m}{5} \right)\| \geq 1$. Done.

$\blacksquare$

PS. If you find error or better exposition leave comment. If you do not want to be in the list of those talking to crackpot trisector, leave a note in the comment, the comment will not appear public, but the content will be implemented in the text.

PPS. There is a much shorter prove for $n=4$ in Barajas and Serra(2008). Algebraically it is short, but I think it lucks an intuition. I have filling this is quite general approach, and cases $n=5$ and $n=6$ bring new ingredients (non prime $n+1$, and competition between 2 pairs of paired runners ). May be we need to wait for the new ingredient until $n=3 \times 5 -1 = 14$.