To get the square of an oriented graph (or any directed graph) you leave the vertex set the same, keep all the arcs, and for each pair of arcs of the form (u,v), (v,w), you add the arc (u,w) if that arc was not already present. An example of an oriented graph and its square is shown above.
Here is the open problem: Prove that for every oriented graph, D, there exists a vertex whose out-degree at least doubles when you square the oriented graph. In the example above, the vertices A, B, C, E and G satisfy this property. (For vertices A and G, 2*0=0). Nate learned of this problem from Paul Seymour. David Fisher proved this theorem for tournaments, i.e., orientations of complete graphs.
Of course, since this is an open problem, no-one can claim to know just what field the problem lies in, since no-one knows the solution.
I believe it is not even known if there are more than 5 points on the parabola which satisfy the condition. It is certainly not known if there is an infinite family with pairwise rational distances. We can quickly see that there are three such points in the following way: Pick two rational numbers, say 1/5 and 3/5, and let those be the distances between the pairs of points AB and BC. If you fix those distances and slide the points up the parabola, the distance AC will gradually increase, bounded above by 4/5. Since the rational numbers are dense in the reals, there will be many placements of the points so that AC is a rational distance.
It is not too hard to show that if you have N points on the parabola with rational distances between them, then you can find N points on the parabola with integer distances between them.