8
$\begingroup$

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 &=& 1 \\ x^2 - 3y^2 &=& 1 \\ x^2 + 6y^2 &=& 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

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

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

$$ \left[ \begin{array}{cc} a + b \sqrt{2} & c - d \sqrt{2}\\ 3(c + d \sqrt{2}) & a - b \sqrt{2} \end{array} \right] = \left[ \begin{array}{cc} 3 + 1 \sqrt{2} & 2 - 1 \sqrt{2}\\ 3(2 + 1 \sqrt{2}) & 3 - 1 \sqrt{2} \end{array} \right]^n $$

$\endgroup$
  • 2
    $\begingroup$ I get a family of solutions described by $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=(3+2\sqrt{2})^m(2+\sqrt{3})^n$ for all integers $m$ and $n$, plus sign-changed equivalents. $\endgroup$ – Oscar Lanzi Sep 7 '18 at 1:26
  • $\begingroup$ $3,1,2,1$ by cursory trial and error, but that doesn't preclude other solutions. $\endgroup$ – Keith Backman Sep 7 '18 at 1:58
  • $\begingroup$ Maybe better to put the problem differently? $$aX^2+bY^2-cZ^2-gQ^2=t$$ $\endgroup$ – individ Sep 7 '18 at 5:38
0
$\begingroup$

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)

$\endgroup$
-1
$\begingroup$

$$x^2-2y^2-3z^2+6q^2=1$$

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

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

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

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

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

$\endgroup$
  • 2
    $\begingroup$ decision of what? $\endgroup$ – cactus314 Sep 7 '18 at 11:42

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.