<s>[INST] 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.
 [/INST] (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,
$$p^4-101^2q^4 = w^2$$
</s>