# When can we solve a diophantine equation with degree $2$ in $3$ unknowns completely?

The diophantine equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ can be solved completely : for every sixtupel $(A,B,C,D,E,F)$ we can determine the complete set of integer pairs satisfying the equation.

What about the more complicated diophantine equation $$Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0$$ ?

In short, a diophantine polynomial equation with degree $2$ in $3$ unknowns. I don't think that we can fully solve such an equation in general, can we ? In which cases can we solve it completely ?

• Why are you doing this? The formula is in General very cumbersome. When turning the numbers is necessary to simplify and change. For the more simple case of what I wrote there. math.stackexchange.com/questions/1513733/… How much of this formula do not write it is removed. So, why to write it? Everyone wants to get short and simple, and it can't be. – individ Sep 26 '17 at 10:22
• @individ Because this is the general equation, and I am interested what we can do in general and when we can solve it. – Peter Sep 26 '17 at 11:58
• The equation in General form can be expressed through the solution of the equation Pell. For the more simple it is possible such to write. math.stackexchange.com/questions/794510/… It is certainly possible to write this formula. Only it is very bulky. Sense to write it if still will not need anyone? – individ Sep 26 '17 at 12:06
• Pell-equation ? We have $3$ unknowns. Can we actually reduce this equation to a pell-equation ? (I only know Pell-equations with $2$ unknowns) – Peter Sep 26 '17 at 12:10
• The formula I showed? math.stackexchange.com/questions/794510/… It's the same 3 unknowns, and reduced to a Pell equation. There's just the option embedded in the coefficient of the equation Pell. – individ Sep 26 '17 at 12:13

I would recommend starting with the simple case $ax^2 + by^2 + cz^2 = d$, since you can usually reduce to this case by completing the square three times. The "parabolic" case $ax^2 + by^2 + cz = d$ should be solvable using modular arithmetic (mod $c$).