How to have this equation $s^2-2(p+q+r+2pqr)s+(p^2+q^2+r^2-2(pq+qr+rp)-4)=0$？ Old Question:
For $x,y,z\in N^{+}$, if such $(xy+1)(yz+1)(zx+1)$ is a perfect square ,show that
$$(xy+1),(yz+1),(xz+1)$$ are all perfect square .
and I konw this PDF have solution, http://math.elinkage.net/attachment.php?aid=18
also this have post solution:http://math.elinkage.net/showthread.php?tid=49&my_post_key=7753a3538fd25446d120928362d64e03&language=chs
and I read this paper and this link.
My question: this key solve this problem  why consider this follow equation
$$s^2-2(p+q+r+2pqr)s+(p^2+q^2+r^2-2(pq+qr+rp)-4)=0$$
How to have this equation?Thank you
 A: The inventor of this solution had the idea to consider $(xy+1)(yz+1)(zy+1)$ as the discriminant of a certain quadratic equation. The question is which equation.
Let's say that it has the form $s^2+bs+c=0$, implying that $b^2-4c=(xy+1)(yz+1)(zy+1)$. Suppose we want to keep the symmetry in $x,y,z$, which means $b$ and $c$ should be symmetric functions of $x,y,z$. Expanding and rewriting, we find
$$(xy+1)(yz+1)(zy+1)=x^2y^2z^2+xyz(x+y+z)+xy+yz+zx+1.$$
The first two terms are of the form $g^2+gh$. In number theory it is often useful to rewrite such expressions as $\frac14(4g^2+4gh)=\frac14((2g+h)^2-h^2)$. So far, we have
$$(xy+1)(yz+1)(zy+1)=\frac14((2xyz+x+y+z)^2-(x+y+z)^2)+xy+yz+zx+1.$$
This can be rewritten as
$$\begin{align*}4(xy+1)(yz+1)(zy+1)
&=(2xyz+x+y+z)^2-(x+y+z)^2+4(xy+yz+zx+1)\\
&=(2xyz+x+y+z)^2-(x^2+y^2+z^2)+2(xy+yz+zx)+4.\end{align*}$$
In order to get something of the form $b^2-4c$ we multiply by $4$:
$$16(xy+1)(yz+1)(zy+1)=(2(2xyz+x+y+z))^2-4(x^2+y^2+z^2-2xy-2yz-2zx-4).$$
This means it is a good idea to consider $b=\pm\,2(2xyz+x+y+z)$ and $c=x^2+y^2+z^2-2xy-2yz-2zx-4$, i.e. the equation
$$s^2\pm\,2(2xyz+x+y+z)s+(x^2+y^2+z^2-2(xy+yz+zx)-4)=0.$$
Note that it has discriminant $16(xy+1)(yz+1)(zy+1)$, but since $16$ is a square this is practically what we intended.
Finally, observe that this is almost symmetric in $x,y,z$ and $s$ too: If we take the $-$ sign, this can be rewritten as
$$s^2+x^2+y^2+z^2-2(xy+yz+zx+sx+sy+sz)-4xyz-4=0.$$
This equation is even more symmetric than we hoped, so certainly pointing towards a solution.
