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Pythagorean triples related to non-isometric equidistant plane quadruples

QUESTION   Do there exist integers   $u\ x\ A\ B$   such that   $x\ne 0$,   and the following two equalities hold:

  • $ x^2 + (x-u)^2\ =\ A^2$
  • $ x^2 + (x+u)^2\ =\ B^2$

?

REMARK   I have a family of pairs of quadruples   $S\ T\subseteq\mathbb Z^2$,   parametrized by   $(u\ x)$,   such that   $S\ T$ have the same six distances but are not isometric (with respect to the Euclidean distance). All six distances of such quadruples are integers   $\Leftrightarrow$   both integers   $x^2+(x-u)^2$   and   $x^2+(x+u)^2$   are full squares.