I am looking for a reference or an elementary proof of the following fact:
The orthogonal group of the lattice $I_{1,2} = I^+ \oplus 2 I^-$ is infinite.
Here the Lorentzian lattice $I_{1,n}$ is given by $\mathbb Z ^{n+1}$ with the quadratic form $\varphi(x) = x_1^2 - \sum_{j=2}^{n+1} x_{j}^2 $. By orthogonal group I mean the group of isomorphisms $f:I_{1,n}\to I_{1,n}$ preserving the quadratic form.
As mentioned, the orthogonal group of the form $I_{1,2}$ that gives Pythagorean triples is infinite. We are asking about square integer matrices $P$ with $P^T H P = H,$ where $$ H = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right) $$
One choice for $P$ is
$$ P = \left( \begin{array}{rrr} 3 & 2 & 2 \\ 2 & 2 & 1 \\ 2 & 1 & 2 \\ \end{array} \right) $$ which is itself symmetric.
$$ \left( \begin{array}{rrr} 3 & 2 & 2 \\ 2 & 2 & 1 \\ 2 & 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 2 & 2 \\ 2 & 2 & 1 \\ 2 & 1 & 2 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right) $$ which you should verify by hand.
I deliberately chose $P$ that had strictly positive elements. The point is that any $P^n$ for positive exponent $n$ is also an orthogonal matrix for $H.$ And the $(1,1)$ entry of $P^n$ grows strictly as well, so these are distinct. Thus the group is infinite. The $(1,1)$ entries $w_n$ for this sequence are $3, 17, 99, 577, ... $ and obey $w_{n+2} = 6 w_{n+1} - w_n$ and give the $w$ value in $w^2 - 2 v^2 = 1.$
The group is infinite but not cyclic. If it were cyclic, all the matrices would commute, but that does not happen. Take a new letter,
$$ Q = \left( \begin{array}{rrr} 9 & 8 & 4 \\ 8 & 7 & 4 \\ 4 & 4 & 1 \\ \end{array} \right) $$
such that $Q^T H Q = H$ Then the new $I_{1,2}$ orthogonal matrices $PQ$ and $QP$ are distinct, still all positive entries, but distinct and not symmetric.
Magnus, on pages 124-125, parametrizes the matrices, two types.
With $a,b,c,d$ all odd and $|ad-bc| = 2,$ we take $$ P = \left( \begin{array}{ccc} \frac{a^2 + b^2 + c^2 + d^2}{4} & \frac{ab+cd}{2} & \frac{a^2 - b^2 + c^2 - d^2}{4} \\ \frac{ac+bd}{2} & \frac{ad+bc}{2} & \frac{ac-bd}{2} \\ \frac{a^2 + b^2 - c^2 - d^2}{4} & \frac{ab-cd}{2} & \frac{a^2 - b^2 - c^2 + d^2}{4}\\ \end{array} \right) $$ For my original matrix $P$ above take $a=3,b=1,c=1,d=1.$
With $a + b + c + d \equiv 0 \pmod 2 $ and $|ad-bc| = 1,$ we take $$ Q = \left( \begin{array}{ccc} \frac{a^2 + b^2 + c^2 + d^2}{2} & ab+cd & \frac{a^2 - b^2 + c^2 - d^2}{2} \\ ac+bd & ad+bc & ac-bd \\ \frac{a^2 + b^2 - c^2 - d^2}{2} & ab-cd & \frac{a^2 - b^2 - c^2 + d^2}{2}\\ \end{array} \right) $$ For my original matrix $Q$ above take $a=3,b=2,c=2,d=1.$