Solve $a^2 - 2b^2 - 3 c^2 + 6 d^2 =1 $ over integers $a,b,c,d \in \mathbb{Z}$ Are we able to completely solve this variant of Pell equation?
$$ x_1^2 - 2x_2^2 - 3x_3^2 + 6x_4^2 = 1 $$
This has an interpretation as is related to the fundamental unit equation of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$ as well as various irreducible quartics.  Is this the same as solving three separate Pell equations?

\begin{eqnarray*}
x^2 - 2y^2 &amp;=&amp; 1 \\
x^2 - 3y^2 &amp;=&amp; 1 \\
x^2 + 6y^2 &amp;=&amp; 1 \tag{$\ast$}
\end{eqnarray*}

Our instinct suggests there should be three degrees of freedom here, and setting different variables to zero we could find three two of these (the third equation has no solutions over $\mathbb{R}$). Does that generate all the solutions?



Perhaps I should remark this quadratic form is also a determinant

$$  a^2 - 2b^2 - 3c^2 + 6d^2 = \det \left[
\begin{array}{cc} 
a + b \sqrt{2} &amp; c - d \sqrt{2}\\
3(c + d \sqrt{2}) &amp; a - b \sqrt{2}
\end{array}
 \right]$$

This might not even contain Oscar's solution.  Extending Keith's solution We could have:

$$
\left[
\begin{array}{cc} 
a + b \sqrt{2} &amp; c - d \sqrt{2}\\
3(c + d \sqrt{2}) &amp; a - b \sqrt{2}
\end{array}
 \right] = 
\left[
\begin{array}{cc} 
3 + 1 \sqrt{2} &amp; 2 - 1 \sqrt{2}\\
3(2 + 1 \sqrt{2}) &amp; 3 - 1 \sqrt{2}
\end{array}
 \right]^n
 $$
Equation given above is shown below:

$a^2-2b^2-3c^2+6d^2=1$

As shown by "Individ" , what "OP" can do is to take (a,b,c,d) as shown below:

$a=p(6w^2+4w+3)$

$b=p(2w^2+6w+1)$

$c=p(4w^2+4w+2)$

$d=p(2w^2+4w+1)$

Where $(p)= [1/(2w^2-1)]$

For suitable value's of 'w' we get the numerical solutions below:

w=(1),      (a,b,c,d)=(13, 9, 10, 7)

w=(3/4),     (a,b,c,d)= (75, 53, 58, 41)

w=(5/7),     (a,b,c,d)= (437, 309, 338, 239)
$$x^2-2y^2-3z^2+6q^2=1$$

Use what any decision  $a^2-2b^2=1$ and $c^2-2d^2-3k^2+6t^2=1$

$$x=ac\pm{2bd}$$

$$y=ad\pm{bc}$$

$$z=ak\pm{2bt}$$

$$q=at\pm{bk}$$