# Solve Diophantine equation with three variables part two

I want to find all solutions of $$x^2+y^2+z^2-xy-yz-zx-x-y-z=0$$ Solutions need not to be primitive. I found several parametric family. For example $$(m^2, m^2+m , (m+1)^2)$$

$$(m^2, m^2+m+2 , (m+1)^2)$$

$$(m^2+1, m^2+m, (m+1)^2+1)$$

$$(m^2+1, m^2+m+4, (m+1)^2+1)$$

$$(m^2-m, m^2, (m+1)^2-(m+1))$$

$$(m^2-m, m^2+1, (m+1)^2-(m+1))$$

$$((m+2)^2-m, (m+2)^2, (m+3)^2-(m+1))$$

$$((m+2)^2-m, (m+2)^2+5, (m+3)^2-(m+1))$$

$$((m+2)^2+5, (m+2)^2+m+4, (m+3)^2+5)$$

$$((m+2)^2+5, (m+2)^2+m+12, (m+3)^2+5)$$

$$((m+3)^2-m+1, (m+3)^2+1, (m+4)^2-m)$$

$$((m+3)^2-m+1, (m+3)^2+8, (m+4)^2-m)$$

$$((m+4)^2+8, (m+4)^2+m+8, (m+5)^2+8)$$

$$((m+4)^2+8, (m+4)^2+m+18, (m+5)^2+8)$$

$$((m+3)^2+m+13, (m+5)^2+5, (m+4)^2+m+14)$$

$$((m+3)^2+m+13, (m+5)^2+16, (m+4)^2+m+14)$$

I suspect I am missing something. These type of single-valued parametric solutions may exist in infinite numbers. Any help will be much appreciated.

• You have one equation with $3$ variables. So I suspect the general solution will be described by a $2$-parameter family. Sep 10 '19 at 6:33
• Yeah, that's my thought. But haven't been able to find 2-parameter family. Sep 10 '19 at 7:09

A bit of abstract theory makes the problem a lot simpler; homogenising the equation into $$X^2+Y^2+Z^2-XY-YZ-ZX-XW-YW-ZW=0,$$ yields a homogeneous equation in four variables of degree $$2$$. This means it is a projective quadric surface in projective $$3$$-space, for which there is a standard method of parametrization if it is nondegenerate and nonempty:

Pick a point $$p$$ on the hypersurface; then every line through $$p$$ intersects the hypersurface in precisely two points (except the tangent line). Conversely every point on the surface of course defines a unique line through $$p$$, except the starting point itself. In this way the points on the hypersurface are parametrized by the lines passing through $$p$$, which are in turn parametrized by the projective plane.

Luckily the given surface is nondegenerate and nonempty; it is defined by the matrix equation $$[X,Y,Z,W]\left[\begin{matrix}2&-1&-1&-1\\-1&2&-1&-1\\-1&-1&2&-1\\-1&-1&-1&0\end{matrix}\right]\left[\begin{matrix}X\\Y\\Z\\W\end{matrix}\right]=0,$$ where the defining $$4\times4$$-matrix is routinely verified to be nonsingular, so the surface is nondegenerate. It is nonempty because we can easily find a point on it by inspection; for example $$(X:Y:Z:W)=(0:0:0:1),$$ corresponding to $$(x,y,z)=(0,0,0)$$ for the original equation. So we can parametrize the surface as described above. But experience shows that the parametrization comes out much more nicely if we choose a point at infinity on the original surface, i.e. a point with $$W=0$$. One such point is $$p:=(X:Y:Z:W)=(1:1:1:0),$$ and from here we can start the dirty work of finding an explicit parametrization.

The lines through the point $$p$$ are parametrized by $$(\lambda:\mu:\nu)\in\Bbb{P}^2$$ as follows: $$L_{(\lambda:\mu:\nu)}:=\{(r+\lambda s:r+\mu s:r-(\lambda+\mu)s:\nu s):(r:s)\in\Bbb{P}^1\}.$$ Every such line intersects the original hypersurface in exactly two points; plugging the parametrization into the equation yields $$\begin{eqnarray*} (r+\lambda s)^2+(r+\mu s)^2+(r-(\lambda+\mu)s)^2&-&(r+\lambda s)(r+\mu s)-(r+\lambda s)(r-(\lambda+\mu)s)-(r-(\lambda+\mu)s)(r+\mu s)\\ &-&(r+\lambda s)(\nu s)-(r+\mu s)(\nu s)-(r-(\lambda+\mu)s)(\nu s)=0, \end{eqnarray*}$$ and simplifying the mess above leaves us with $$s(\nu r-(\lambda^2+\lambda\mu+\mu^2)s)=0,$$ yielding the nontrivial solutions $$(r:s)=(\lambda^2+\lambda\mu+\mu^2:\nu),$$ corresponding to the intersection point $$(X:Y:Z:W)=(\lambda^2+\lambda\mu+\mu^2+\lambda\nu:\lambda^2+\lambda\mu+\mu^2+\mu\nu:\lambda^2+\lambda\mu+\mu^2-\lambda\nu-\mu\nu:\nu^2).$$ By setting $$\nu=1$$ this reduces to the affine parametrization $$(x,y,z)=(l^2+lm+m^2+l,l^2+lm+m^2+m,l^2+lm+m^2-l-m),$$ with parameters $$l$$ and $$m$$.

I was wrong about the formula... it looks like this....

$$X^2+Y^2+Z^2=XY+XZ+ZY+X+Y+Z$$

$$X=s(3(k^2-kt+t^2)s-k-t)$$

$$Y=s(3(k^2-kt+t^2)s+2k-t)$$

$$Z=s(3(k^2-kt+t^2)s-k+2t)$$

• If you want this to be answer you have to explain a little more ... Sep 10 '19 at 7:08
• Does this parametric family generate all the solutions? And $k,s,t$ take any positive or negative integer value? Sep 10 '19 at 7:08
• It could be anyone. Both positive and negative numbers. Sep 10 '19 at 7:13
• How do we know that this gives all solutions? Sep 10 '19 at 8:17