In 1643, Fermat asked Frenicle et al to find a special Pythagorean triple $a,b,c$ such that for $n=1$,
$$a+nb = r_1^2\\ a^2+b^2 = r_2^4\tag1$$
Equivalently,
$$\color{blue}{\big((p^2-q^2)^2-(2pq)^2\big)}+n\color{brown}{\big(2(p^2-q^2)(2pq)\big)} = r_1^2\tag2$$
This has polynomial solution $p,q = n^2+2, 2n.$ (And an infinite more, since it can be birationally transformed to an elliptic curve.) But this simple solution yields negative $a$ for $n=1,2,3,4$.
Fermat stated that the smallest positive answer for $n=1$ was,
$$n,a,b = 1,\;4565486027761,\; 1061652293520$$
Q: What is the smallest positive solution for $n=2,3,4$?
For $n=2$, I think it is,
$$n,a,b = 2,\;10386304269597791463121,\; 3672193546323671330640$$
though I am not sure.