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Could you please help with finding of general solution of diophantine system for rational a, b, c, d

$(a^2+b^2)(c^2+d^2)=A^2$

$(a^2-b^2)(c^2-d^2)=B^2$

for some rational A and B.

This is related to Ozanam's problem.

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  • $\begingroup$ I think the field of solutions of this system should be similiar to $(a^4-b^4)(c^4-d^4)=X^2$ does this make sense? $\endgroup$
    – veg_nw
    Commented Apr 14, 2016 at 21:06
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    $\begingroup$ Not necessarily. A solution to $(a^4-b^4)(c^4-d^4) = x^2$ is $a,b,c,d =1,\,3,\,31,\,49$, but it does not solve $ (a^2\pm b^2) (c^2\pm d^2) = \square$. $\endgroup$
    – Tito Piezas III
    Commented Apr 16, 2016 at 1:07
  • $\begingroup$ Yes. Sure. Obviously the set of solutions of $(a^4-b^4)(c^4-d^4)=X^2$ is larger. Would be interesting to get the general solution for this equation also. May be we can filter out the solutions of $(a^2\pm b^2) (c^2\pm d^2) = \square$ from there. $\endgroup$
    – veg_nw
    Commented Apr 16, 2016 at 6:30
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    $\begingroup$ And also I have found the following article with some set of solutions, but not sure if this is a general solution. ac.els-cdn.com/S0315086084710056/… page 15. $\endgroup$
    – veg_nw
    Commented Apr 16, 2016 at 6:31
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    $\begingroup$ Actally I am interested in only $S_1$ because of the task I am currently working on. From this paper I have found the following solutions: $c=-{k}{a}( {a}^{8}+6\,{a}^{4}{b}^{4}-3\,{b}^{8}); d=-{k}{b}( 3\,{a}^{8}-6\,{a}^{4}{b}^{4}-{b}^{8}) $. But not sure if this is the general solution of this system. $\endgroup$
    – veg_nw
    Commented Apr 16, 2016 at 22:23

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