On $p^2 + nq^2 = z^2,\;p^2 - nq^2 = t^2$ and the "congruent number problem"

Question

(Much revised for brevity.) An integer $n$ is a congruent number if there are rationals $a,b,c$ such that,

$$a^2+b^2 = c^2\\ \tfrac{1}{2}ab = n$$

or, alternatively, the elliptic curve,

$$y = x^3-n^2x = x(x-n)(x+n)\tag1$$

is solvable in the rationals. Assume $x=(p/q)^2$. Then $(1)$ becomes,

$$\frac{p^2}{q^6}(p^4-n^2q^4) = y^2$$

or simply,

$$(p^2+nq^2)(p^2-nq^2) = w^2\tag2$$

Assuming $w=z\,t$ and equating factors, then $(2)$ becomes,

$$p^2 + nq^2 = z^2\\ p^2 - nq^2 = t^2\tag3$$

This is the implication given by Mathworld and OEIS.

Questions:

1. Is it true that if $p^4-n^2q^4 = w^2$ is solvable, then is $$p^2 + nq^2 = m z_1^2\\ p^2 - nq^2 = m z_2^2\tag4$$ necessarily also solvable for $m=1$?
2. Is the solution to $(4)$ for $m=1$ the smallest for $p^4-n^2q^4 = w^2$?

P.S. This post made me re-visit congruent numbers.

Answer 1

(Edited 2018): After 2 years finally found the answer.

I. For Question $1$, the answer is yes.

If one has a solution to $x^4-n^2y^4 = z^2$, then,

$$p^2-nq^2 = (z^2-2nx^2y^2)^2\\ p^2+nq^2 = (z^2+2nx^2y^2)^2$$

where $p=x^4+n^2y^2$ and $q=2xyz$.

II. For Question $2$, the answer is no.

For example, in this Mathworld link, we see that the smallest solution to,

$$p^2+101q^2 = \color{brown}mz_1^2\\p^2-101q^2 = \color{brown}mz_2^2$$

for $ \color{brown}{m=1}$ is the large,

$$p_1,\,q_1 =2015242462949760001961,\, 118171431852779451900\tag1$$

but for $\color{brown}{m=101}$ is the smaller,

$$p_2,\,q_2 =2125141,\, 63050\tag2$$

Thus, $(2)$ is a smaller solution to,

$$p^4-101^2q^4 = w^2$$

Answer 2

Probably for the system.

$$\left\{\begin{aligned}&x^2+qy^2=a^2\\&x^2-qy^2=b^2\end{aligned}\right.$$

Write a simple formula as these numbers just to find.

$$x=p^4+s^4$$

$$y=2ps\sqrt{\frac{(p^2+s^2)(p^2-s^2)}{q}}$$

$$a=p^4+2p^2s^2-s^4$$

$$b=s^4+2p^2s^2-p^4$$

Or so.

$$x=p^4+6p^2s^2+s^4$$

$$y=2(p^2-s^2)\sqrt{\frac{2ps(p^2+s^2)}{q}}$$

$$a=p^4+4sp^3-2p^2s^2+4ps^3+s^4$$

$$b=p^4-4sp^3-2p^2s^2-4ps^3+s^4$$

It is clear that the number $q$ must be chosen so that the root was rational. Of course it's not really a solution, but it can generate and try all possible options. For finding solutions to the desired number - the task can be facilitated by considering the possible options - factorization.