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



*

*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 $m=1$? 

*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.

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.
