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.

It's not even possible for the product of $x^2 + (x-u)^2$ and $x^2 + (x+u)^2$ to be a square unless $x=0$ or $u=0$: that product is $4x^4+u^4$, and
the elliptic curve $4x^4+u^4 = y^2$ is isomorphic to $Y^2 = X^3 - X$, 
for which Fermat already proved that the obvious rational points are 
the only ones.  For your application $x=0$ is forbidden and $u=0$ makes 
$A^2 = B^2 = 2x^2$ which would again imply $x=0$.
To get from $4x^4+u^4=y^2$ to $Y^2 = X^3 - X$, let $y = 2x^2-m$ to get
$4mx^2 = m^2-u^4$.  If $x \neq 0$ this makes $m(m^2-u^4)$ a square.
Then if $u\neq 0$ then write $m = Xu^2$ to get $u^6(X^3-X)$, etc.
