# General Quadratic Diophantine Equations of Three Variables

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.

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 $$\mathbf Q$$.