Unsolved problems

The conjecture about the lonely runner


Some athletes start together at the same starting point for a running contest over a circular track. They keep running clockwise around forever at fixed distinct speeds. At some instant of time each runner may be at equal distance d from the one who runs before him and the one behind him, so that the positions of the runners are evenly distributed over the track. But is there for each of them an instant of time when he is at distance at least d from any other runner?
The conjecture that JM Wills made in 1967 says there is such an instant of time. The problem is how to prove this. This problem has been solved for up to seven runners, but the general problem remains unsolved.

In mathematical terms, we may formulate the problem for k+1 runners as follows:
Given a positive integer k and k integers d1 , d2 , ... , dk without common prime divisor. Prove there is a positive real t such that for all i∈{1,2,..,k} the product tdi is at distance at least 1/(k+1) from the nearest integer.

First approach: A computer program

Denote the distance of any real number x from the nearest integer by ||x||.

We may use the following Pascal program to search for an approximation of the minimal t if k is at most 100 and the k given integers are within standard pascal range:

program unsolved;
var i,j,k:integer; t:real; d:array[1..100] of integer; noggoed:boolean;
function norm(x:real):real;
  writeln('k?'); readln(k);
  writeln('Geef ',k:3,' positieve integers');
  for j:=1 to k do begin writeln('d[',j:3,']?'); readln(d[j]) end;
    noggoed:=true; i:=0;
    while ((i < k) and noggoed) do
      begin i:=i+1; if norm(t*d[i]) < (1/(k+1)) then noggoed:=false end;
    if noggoed then
        writeln(t:13:5); for j:=1 to k do writeln(norm(t*d[j]):13:10);
        writeln; readln
  until noggoed

The 3n+1 conjecture


Take an arbitrary positive integer n. If n is even, change it into n/2. If n is odd, change it into 3n+1. Show that if we repeat this procedure, we get 1 in the end.

First approach:

If n is even, then the probability that n/2 is even is 1/2 again. We count n → n/2 as one step.
If n is odd, then 3n+1 is even and the probability that (3n+1)/2 is even is 1/2 again. We count n → 3n+1 → (3n+1)/2 as one step.
If we neglect the +1 in (3n+1)/2 then after k steps we expect to have n*3k/2/2k = n*(√3/2)k, which tends to 0 if k → ∞ unless we get 1 at some point.
So we should get 1 in the end.

The square in the loop


Prove that in any loop that doesn't intersect itself there exists an inscribed square.

First approach:

With a pascal program we may systematically search for four points on the loop that are the vertices of a square.
As an example, I did this with the raindrop (t3-3t,3t2), -√3 ≤ t ≤ √3, and found (approximated values of t, x, y, resp):

A -1.28 -1.74 4.93
B -0.69 -1.75 1.44
C 0.69 1.75 1.44
D 1.28 1.74 4.93