Is it true that $n = 2t^2-2$ for even $t$ is a congruent number? This post asks for $m$ such that the simultaneous Pythagorean triples,
$$a^2+m^2b^2 = c^2\\b^2+c^2 = d^2\tag1$$
have solutions. Will Jagy found an infinite family given by,
$$m = 2t^2-2 = 0, 6, 16, 30, 48, 70, 96, 126, \dots$$
where,
$$\begin{aligned}
a &= -1 + 9 t^2 - 12 t^4 + 4 t^6\\ b &=  -2 t + 4 t^3\\c &=  -1 + t^2 - 4 t^4 + 4 t^6\\ d  &=  1 + t^2 - 4 t^4 + 4 t^6
\end{aligned}$$

The values $m=6,30,70$ were faintly familiar, as I had posted about congruent numbers before. (See this and this.) A number $n$ is congruent if there is a solution to the simultaneous,
$$p^2 + nq^2 = r^2\\
p^2 - nq^2 = s^2\tag2$$ 

Q1: Is it true that an infinite family of congruent numbers is given by,
  $$n = 2(2v)^2-2 = 6,30,70,126,\dots$$

P.S. A003273 gives a list of congruent numbers $N<10000$ and all $n$ of that form are there.

Q2: If indeed true, what is the connection between systems $(1)$ and $(2)$? 


(This is a partial answer.)
After some persistence and effort, I managed to find a partial answer. It can be proven that
$$n = 2t^2-2$$
is a congruent number for infinitely many $t$ (odd or even).
Proof: If,
$$n = 2(v^2\pm3)^2-2$$
then,
$$p^2+nq^2=r^2\\p^2-nq^2 = s^2$$
has the simple solution,
$$\begin{aligned}p &=v^4\pm4v^2+8
\\q &=2v\end{aligned}$$ 
