0
$\begingroup$

For a quadratic diophantine equation of two variables, $Ax^2+Bxy+Cy^2 =D$, it's not difficult to find the solutions as it is a generalized Pell equation. However, what happens when we incorporate more variables? Is there any information on the diophantine equation of three variables $Ax^2+By^2 + Cz^2 + Dxz + Exy + Fyz + Gxyz = h$? Any information or references would be appreciated.

$\endgroup$
0
$\begingroup$

To avoid misunderstanding, let me recap your question : you want to study the general quadratic diophantine equation (i.e. with solutions in $\mathbf Z$) in $3$ variables $f(x,y,z)=h$. My suggestion is to homogenize it on mutiplying $h$ by $t^2$ and study instead the homogeneous rational (i.e. with solutions in $\mathbf Q$) in $4$ variables $g(x,y,z,t)=0$. This will give necessary conditions for the original equation.

In the language of quadratic forms, it is said that the rational form $g(x,y,z,t)$ represents $0$ if there exists a non zero quadruple s.t. $g(x,y,z,t)=0$. The existence problem is entirely solved by the celebrated global-local Hasse-Minkowski theorem : Let $K$ be a number field and $G$ a non degenerate quadratic form in $n$ variables. Then $G$ represents $0$ in $K$ iff it represents $0$ in all the completed fields $K_v$ for all valuations $v$ of $K$ (archimedean or not). More concretely : 1) If $n\ge5$, then $G$ represents $0$ unless there is a $v$ s.t. $K_v=\mathbf R$ ; 2) The case $n=4$ can be brought back to $n=3$ ; 3) The cases $n=1,2$ are trivial. So the crucial remaining cases are $n=3,4$ . If $n=3$, we can diagonalize our form as $G=x^2 - by^2-cz^2$, and the existence criterion then reads : $G$ represents $0$ in $K$ iff $c$ is a norm from $K(\sqrt b)$ (this is purely algebraic), iff all the local quadratic norm residue symbols $(b,c)_v$ are trivial (this is CFT). If $n=4$ and $G=x^2-by^2-cz^2+act^2$, then $G$ represents $0$ in $K$ iff $x^2 - by^2-cz^2$ represents $0$ in $K(\sqrt {ab})$(purely algebraic). For all these assertions I refer to Cassels-Fröhlich, ANT, exercise 4. In the particular case $K=\mathbf Q$, the criterion for $n=3,4$ can be made more precise : let $n=3$, or $n=4$ and the dicriminant of $G$ is not in $\mathbf {Q^*}^{2}$; if $G$ represents $0$ in all $\mathbf Q_v$ except at most one, then $G$ represents $0$ in $\mathbf Q$.

Note that the above discussion gives only the existence of solutions, not their explicitation, even over $\mathbf Q$.
NB. Concerning your additional query "Does anyone have any thoughts on this? /bump", I have no answer ./.

$\endgroup$

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.