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,

$$m = 2t^2-2 = 0, 6, 16, 30, 48, 70, 96, 126, \dots$$


$$\begin{aligned} a &= -1 + 9 t^2 - 12 t^4 + 4 t^6\\ b &= -2 t + 4 t^3\\c &= -1 + t^2 - 4 t^4 + 4 t^6\\ d &= 1 + t^2 - 4 t^4 + 4 t^6 \end{aligned}$$

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,

$$n = 2(v^2\pm3)^2-2$$


$$p^2+nq^2=r^2\\p^2-nq^2 = s^2$$

has the simple solution,

$$\begin{aligned}p &=v^4\pm4v^2+8 \\q &=2v\end{aligned}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.