# 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.

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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.
If I'm correct, Fermat's proof by descent that $x^4+y^4=z^2$ has no solutions can be adapted to the equation $4x^4+u^4=y^2$. It's just a matter of paying careful attention to the 2's. – Barry Cipra Jun 27 '13 at 21:48
Wlodzimierz Holsztynski: You're welcome! As it happens this was the second time this week that I was asked about a Diophantine equation that turned out to reduce to this Fermat curve... @Barry Cipra: you're right, and Fermat already did it! He proved that neither $x^4+y^4=z^2$ nor $x^4+y^2=z^4$ has nontrivial rational solutions. The former curve is isomorphic with $y^2=x^3-4x$ and isogenous with $y^2=x^3+x$; the latter, isomorphic with $y^2=x^3+4x$ and isogenous with $y^2=x^3-x$. – Noam D. Elkies Jul 1 '13 at 3:23