Rational distances from the vertices of a square

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Put $A=(0,0)$, $B=(1,0)$, $C=(1,1)$ and $D=(0,1)$ in the Euclidean plane. For $P=(x,y)\in\mathbb R^2$, the four distances are $\sqrt{x^2+y^2}$, $\sqrt{(x-1)^2+y^2}$, $\sqrt{(x-1)^2+(y-1)^2}$ and $\sqrt{x^2+(y-1)^2}$. No restriction to the interior of the square is imposed.

Statement recorded in the supplied source.

Does there exist $P\in\mathbb R^2$ for which all four displayed distances lie in $\mathbb Q$? Equivalently, does there exist a square of positive integer side length in a plane, together with a coplanar point whose distances to all four vertices are integers? Exhibit such a configuration or prove it impossible. [S2]

Short English statement

Can a point anywhere in the plane have rational distances from all four corners of a unit square? Equivalently, can all five lengths become integers after scaling?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 9300004 (AMR-092-0004). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Tim S. Roberts, Unsolved problems in number theory, logic, and cryptography RationalDistance problem statement. https://unsolvedproblems.org/index_files/RationalDistance.htm.