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.


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 .

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One Response to On the lonely runner conjecture III

  1. Tim Roberts says:

    There is a prize of $1000 on offer for a valid solution to the Lonely Runner conjecture. For details see the Unsolved Problems web site at

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