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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 ?

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  • $\begingroup$ 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. $\endgroup$ – individ Sep 26 '17 at 10:22
  • $\begingroup$ @individ Because this is the general equation, and I am interested what we can do in general and when we can solve it. $\endgroup$ – Peter Sep 26 '17 at 11:58
  • $\begingroup$ 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? $\endgroup$ – individ Sep 26 '17 at 12:06
  • $\begingroup$ Pell-equation ? We have $3$ unknowns. Can we actually reduce this equation to a pell-equation ? (I only know Pell-equations with $2$ unknowns) $\endgroup$ – Peter Sep 26 '17 at 12:10
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    $\begingroup$ 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. $\endgroup$ – individ Sep 26 '17 at 12:13
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There are so many cases that it would be hard to list all cases in which we can solve it completely. For example, all of the two-variable quadratic Diophantine equations are special cases, and there are long papers written about various cases of two-variable quadratic Diophantine equations. (For example, John Robertson's page, http://www.jpr2718.org/, has some very nice papers on solving second-degree Diophantine equations in two variables.)

I think that the general answer for the three-variable case is, for most choices of coefficients, there are infinitely many solutions, but there is no neat tidy formula that describes all such solutions. (Of course, in certain special cases, there would be.)

Some simple cases are mentioned here: http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html

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 $c$).

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