This seemingly trivial problem has been apparently still unsolved [3]. If a point P is set in a plane's irrational place (be it inside or outside the square), then at least one of the four distances P to vertices must be irrational. If the point P is inside the square and set in a plane's rational place and all four P to vertices distances are assumed rational, then these distances form hypotenuses of Pythagorean triangles. The distances are - at the same time - hypotenuses of other triangles: triangles formed by irrational legs which are "compatible" with the diagonals of the square (and, of course, not measurable with rational units). Calculation shows that these hypotenuses, if assumed rational, must be all even integers. Since primitive Pythagorean triangles must have odd hypotenuses [1], those triangles are not primitive and should be simplified by division by two. After the first (and all subsequent) divisions the situation doesn't change, the hypotenuses remain even integers and thus divisible by two. That infinite divisibility can be considered as reductio ad absurdum - - a kind of a proof of infinite descent introduced by Fermat [2]. For the point on the border the proof is rather trivial; for the (rationally set) point outside the square other sets of triangles are used to disprove by infinite descent the assumption that the distances can be all rational.