Let $n$ distinct points in the four-dimensional cube $$\{ (x,y,z,t):0\le x, x\le 1,0\le y, y\le 1,0\le z, z\le 1,0\le t, t \le 1 \} $$ be such that all the distances between these points are (strictly) greater than one. How much can $n$ be? I know the answer in three dimensions is 7. Here and here is my Mathematica code which finds 7 points in the 3D cube, producing
{1., {Indexed[x, {1}] -> 1., Indexed[x, {2}] -> 0.736543, Indexed[x, {3}] -> 0.000953636, Indexed[x, {4}] -> 0.999888, Indexed[x, {5}] -> 0.0010957, Indexed[x, {6}] -> 0., Indexed[x, {7}] -> 1., Indexed[y, {1}] -> 0.967476, Indexed[y, {2}] -> 0.00179236, Indexed[y, {3}] -> 0.712727, Indexed[y, {4}] -> 7.67595*10^-8, Indexed[y, {5}] -> 0.000973333, Indexed[y, {6}] -> 0.974891, Indexed[y, {7}] -> 1., Indexed[z, {1}] -> 0.0000193274, Indexed[z, {2}] -> 0.00039306, Indexed[z, {3}] -> 2.28036*10^-7, Indexed[z, {4}] -> 0.965628, Indexed[z, {5}] -> 0.703141, Indexed[z, {6}] -> 0.980439, Indexed[z, {7}] -> 0.99999}}
ListPointPlot3D[Table[{Indexed[x, j], Indexed[y, j], Indexed[z, j]}, {j, 1,
7}] /. %[[2]]]
Eight points cannot be located in such a way (see this article). What is such $n$ in four dimensions?