(This complements the post on $x^4+m^2y^4=z^2$.)
A subset of solutions to, $$(x^2-101y^2)(x^2+101y^2)=z_1^2\tag1$$ also solve, $$x^2-101y^2=z_2^2\tag2$$ $$x^2+101y^2=z_3^2\tag3$$ For example, we have $x,y$ only for eq. $(1)$ $$x=2125141,\;y=63050,\;z_1=4498341355119$$ But since $n=101$ is a congruent number, all three can be solved, the smallest being,
$$x = 2015242462949760001961\\y = 118171431852779451900$$
Q: In general, if for some $n$ the Diophantine $$(x^2-ny^2)(x^2+ny^2)=z_1^2$$ has solutions (infinitely many since an elliptic curve is involved), does that necessarily mean a subset solve the simultaneous $$x^2-ny^2=z_2^2\\ x^2+ny^2=z_3^2$$ as well?