General Quadratic Diophantine Equations of Three Variables

Question

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.

Answer 1

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 ./.