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(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?

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Yes: The pair of equations $x^2 - n y^2 = z_2^2$, $x^2 + n y^2 = z_3^2$ defines a genus 1 curve in $\mathbb P^3$. Since it has obvious rational points (with $y = 0$), it is an elliptic curve, and the map to the original curve $x^4 - n^2 y^4 = z_1^2$ given by setting $z_1 = z_2 z_3$ is a 2-isogeny (when choosing the origins in a compatible way). So the groups of rational points on the two elliptic curves have the same rank; in particular, if one is infinite, then so is the other.

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