Given a unit square lying on XY plane with center at the origin and we search the whole of 3D space, will we able to find points that are at rational distances (Euclidean metric) from all 4 vertices of the square? If "yes", are there infinitely many such points in 3D space?
I understand, the basic (and still open) form of the question asks for a point on the XY plane that is at rational distances from all 4 vertices.
Note: one could just push the envelope a bit and ask if we search in a higher dimensional (>3) space in which the unit square lies on the x y plane, whether there are points that are at rational euclidean distances from all 4 vertices.
The corners of the square, as points in space, are at $$(x,y,z)=(\pm \frac12, \pm \frac12,0)$$ Consider the points of form $$P_n=(x,y,z)=(0,0,\sqrt{n^2-\frac12})$$ Then $P_n$ is at distance $n$ from each corner of the square.
Added later: OP has asked for such points not on the $z$ axis. The following construction gives an infinite collection of points which lie directly over the midpoint of a side of the square. We describe them in terms of the square with corners $(\pm a,\pm a)$ where $a$ is to be chosen later. One could rescale to the desired unit square centered at the origin.
To begin with, choose any Pythagorean triple of form $(2a,b,c)$ for which $a>b.$ Then define $P$ as the point $(a,0,\sqrt{b^2-a^2})$ One can check that the distance from P to either square vertex $(+a,\pm a,0)$ is $b,$ while the distance from $P$ to the other square vertices $(-a,\pm a,0)$ is $c.$