# 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 $(2)$?

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}