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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.

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

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

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

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

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

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

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

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