How to solve $a^2 - 2b^2 - 3 c^2 + 6d^2 = 1$ over $\mathbb{Z}$? I would like to find the integral elements int the quaternion algebra $D_{2,3}(\mathbb{Q})$.  These are $2 \times 2$ matrices with elements in $\mathbb{Z}[\sqrt{2}]$ with determinant $1$.  Observe these matrices are closed under multiplication.
$$ \det \left| \begin{array}{cc} x_1 + \sqrt{2} x_2 & x_3 + \sqrt{2} x_4 \\
3(x_3 - \sqrt{2} x_4) & x_1 - \sqrt{2} x_2\end{array}  \right| = x_1^2 - 2x_2^2 - 3x_3^2 + 6 x_4^2 = 1 $$
Books as recent as 2003, such on John Conway's On Quaternions and Octonions they collect fact about integral quaternions.
The usual mistake here is that I could have written an equation with no solutions, though I suspect that's unlikely.  This is similar to Pell's equation and we're looking for integer points on a conic section, a hyperboloid.

By exhaustive search (a standard problem-solving heuristic) I was able to show that:
$$ \det \left| \begin{array}{cc} 4 + 3 \sqrt{2} & 7 + 5 \sqrt{2}
 \\
3(7-5\sqrt{2}) & 4 - 3 \sqrt{2} 
\end{array}\right| = 4^2 - 2 \times 3^2 - 3 \times 7^2 + 6 \times 5^2 = 1$$
These number theory problems are notorious for their brevity. In fact, they will get arbitrarily complex.  We also get an element of $D_{2,3}(\mathbb{Z})$.
There are books or papers on integer solutions to quadratic forms.  Cassels' book proves outlines and theory of qudratic forms over $\mathbb{Q}$.  For example,  there's no complete list and even writing down generators of this group could be complicated.
 A: You are looking for units of the order $\Lambda$ formed by the matrices, denote them $M(x_1,x_2,x_3,x_4)$ with $x_i,i=1,2,3,4,$ ranging over $\Bbb{Z}$. Obviously Conway can say more about this than I can dream of ever learning. I just wanted to make a comment explaining, why I think this is difficult. But that comment is way too long, so an answer it is.
I once had a need to find the units of a maximal order of another division algebra, but a simple computer search proved that the structure of the group is more complicated than in the case of maximal orders of number fields, and I gave up on the idea. A colleague searched for more information. A number of general facts are known, but IIRC in general the answer is rather messy.
Furthermore, I have not checked whether $\Lambda$ is a maximal order of $D$. That may not be relevant you (and may not have too much of an impact on the answer), so let's ignore that question.
The main comment I want to make is that there are copies of orders of infinitely many quadratic number fields $K, [K:\Bbb{Q}]=2$ inside $\Lambda$. Consider the following.
Any matrix $A=M(x_1,x_2,x_3,x_4)$ satisfies its characteristic polynomial $\chi_A(T)$. The coefficients of that polynomial are the reduced trace and norm of the element $A$ respectively. Because we are in an order we can conclude that $\chi_A(T)\in\Bbb{Z}[T]$.
Furthermore, unless $A$ is in the center (= $\Bbb{Q}$), $\chi_A(T)$ cannot have rational roots. For if $\chi_A(q)=0$, $q\in\Bbb{Q}$, then $qI_2-A$ would be non-invertible contradicting the fact we have a division algebra.
It follows that the set
$$
\Bbb{Q}(A)=\{q_1+q_2A\mid q_1,q_2\in \Bbb{Q}\}
$$
is a splitting field of $\chi_A(T)$, call it $K_A$. That is, it is isomorphic to a quadratic number field. Furthermore, the set
$$\Bbb{Z}[A]=\{n_1+n_2A\mid n_1,n_2\in\Bbb{Z}\}$$
is an order of $K_A$.
By basic facts about quadratic number fields we know that whenever $K_A$ is real (or, equivalently, the discriminant of $\chi_A(T)$ is positive), the order $\Bbb{Z}[A]$
has infinitely many units.
A few examples:
$$
A=M(0,1,1,1)=\left(
\begin{array}{cc}
 \sqrt{2} & 1+\sqrt{2} \\
 3-3 \sqrt{2} & -\sqrt{2} \\
\end{array}
\right)$$
has $\chi_A(T)=T^2+1$, implying that $A$ is itself a unit of order four and $K_A\simeq \Bbb{Q}(i)$. Observe that
$$
-2\cdot1^2-3\cdot1^2+6\cdot1^2=1.
$$
If, instead, we look at the element
$$
A=M(1,1,2,0)=\left(
\begin{array}{cc}
 1+\sqrt{2} & 2 \\
 6 & 1-\sqrt{2} \\
\end{array}
\right)
$$
we get $\chi_A(T)=T^2-2T-13$. The zeros of that polynomial are $1\pm\sqrt{14}$ implying that
$\Bbb{Z}[A]\simeq \Bbb{Z}[\sqrt{14}]$. In that ring $15+4\sqrt{14}$ is a unit (solve a Pell equation to find it). The isomorphism has $A=1+\sqrt{14}$, so
$$
15+4\sqrt{14}=11I_2+4A=M(15,4,8,0)
$$
should also be a unit.
Indeed,
$$
15^2-2\cdot 4^2-3\cdot8^2=1.
$$

Further facts/observations/complications:

*

*There are several quadratic number fields inside your division algebra. By the so called double centralizer theorem each $K_A$ is its own centralizer in $D$. So if $u\notin K_A$ is a unit of $\Lambda$, then the conjugate subfield $uK_Au^{-1}$ is isomorphic to $K_A$ but intersects it only at $\Bbb{Q}$.

*But there exists a method for describing exactly which quadratic number fields occur as (maximal) subfields of $d$. That high powered theory may be overkill for our purposes given that we easily verify
$$
\chi_{M(a,b,c,d)}(T)=T^2-2aT+(a^2-2b^2-3c^2+6d^2)
$$
(should have spelled this out earlier).
And the discriminant of that polynomial is
$$
\Delta=4(2b^2+3c^2-6d^2)
$$
implying that $K_A\simeq\Bbb{Q}(\sqrt{2b^2+3c^2-6d^2})$. Each and every such quadratic field produces its share of units to $\Lambda$. Together with the conjugates.

*Skolem-Noether -theorem tells us that whenever $K_A$ and $K_{A'}$ are isomorphic fields, they are conjugates in the division algebra $D$. Meaning that the units of a given quadratic number field appear as units of $\Lambda$ in many distinct copies gotten by conjugating with other units.

*Nothing I can say will make my claim that the group of units is "messy" more explicit. Hopefully a more knowledgable user shows up and can shed additional light to your question.

A: From an elementary perspective:
Note that the equation is the same as $$a^2-2b^2-3(c^2-2d^2)=1.$$
For nonzero integers $p,q,r,s,t,u,v$:
Let $a=up^2+vq^2$, $b=2pqt$, $c=ur^2+vs^2$, $d=2rst$. Now let $2t^2=uv$. Then the equation is the same as $$(up^2-vq^2)^2-3(ur^2-vs^2)^2=1$$ which reduces to the form $m^2-3n^2=1$ for integers $m$ and $n$. If we let $f(m,n)=m^2-3n^2$ then we can find infinitely many solution pairs $(m,n)$ through $$f(m,n)=f(2m+3n, m+2n)=f(7m+12n, 4m+7n)=\cdots$$ Not sure if this counts as "solving" the equation, but it may help.
