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.
This is a really nice question! I think your answer for $n=2$ is correct.
Equation $(2)$ can be shown to be equivalent to the elliptic curve \begin{equation*} v^2=u^3-(n^2+1)u \end{equation*} with \begin{equation*} \frac{p}{q}=\frac{nu-v}{n^2+1-u} \end{equation*}
The elliptic curve has one finite torsion point at $(0,0)$, though I have not attempted to prove this. There is, also, a point when $u=-1$ and $v=n$, so that the curve always has rank greater than zero.
I wrote a Pari-gp program which uses Denis Simon's ellrank code to determine the rank and generators of the curve. It then searched through rational points to find a minimum positive solution. The results are
n = 1
Rank of curve = 1
Minimum = [4565486027761, 1061652293520]
n = 2
Rank of curve = 1
Minimum = [10386304269597791463121, 3672193546323671330640]
n = 3
Rank of curve = 1
Minimum = [166911677107794033180761521, 383496393351054937416817200]
n = 4
Rank of curve = 2
Minimum = [230903401, 2224677000]
n = 5
Rank of curve = 1
Minimum = [104041, 679320]
n = 6
Rank of curve = 1
Minimum = [53641, 148200]
n = 7
Rank of curve = 1
Minimum = [3744841, 6868680]
n = 8
Rank of curve = 2
Minimum = [876353497971071281, 193103317517647440]
n = 9
Rank of curve = 3
Minimum = [152401, 142800]