<s>[INST] Diophantine question This came up when I did a brand-new (or maybe it's just "birdtracks" in disguise :-)
graph-based construction of the E8 family. x,y,z are dimensions and thus integer. x<0 actually doesn't hurt. (For $y=x,z=(3 (-2 + o) o)/(10 + o)$ the standard E8 setup results, o must now divide 360 etc. pp.) So here is the equation:
$3 x (2 + x) (-2 + x + x^2 - 2 y) y (-x + x^2 - 2 z) z=q^2$       
With rational y or z I wouldn't pester MO - I'd solve it on the spot. But here some nasty division properties are involved, and I'm lousy in number theory. Is there a finite, easy describable solution list?
 [/INST] Let $P(x,y,z)=3 x (2 + x) (-2 + x + x^2 - 2 y) y (-x + x^2 - 2 z) z$ be the product which we wish to be a square. As noted, 


*

*$P(1,y,z)=\left(6yz\right)^2.$


A nice parametric solution is 


*

*$P(2r,3r,r+1)=\left(12r(r+1)(2r^2-2r-1)\right)^2$ 


Also, for fixed $x,y$ or $x,z$ we are left with a Pell Equation.

The parametric solution arises  from first setting $y=x+z-1$ to make the two quadratic factors equal $P(x,x+z-1,z)=(x^2-x-2z)^23x(x+2)z(x+z-1).$  There are many other parametric solutions such as 


*

*$P(r^2,r^2+2,3)=\left(3r(r^2+2)^2(r^2-3)\right)^2$ and 

*$P(3r^2-2,3r^2-2,1)=\left(3r(3r^2-4)(3r^2-2)(3r^2-1)\right)^2.$


Although none that I know of with the right-hand side the square of a fourth degree polynomial in $r$.
Most solutions to $P=\Box$ do not have $y=x+z-1$. There are $1442$ choices of $2 \le,x \le 2000,1 \le z \le 2000$ which make $P(x,x+z-1,z)$ a square. Of them $1000$ have $x=2(z-1).$ 
For fixed $x$ and $z \gt \frac{x^2-x}2$ we have $P(x,y,z)=Ay(y-B)$ for some constants. If $A \gt 0$ has square-free part $\alpha \gt 1$ then we are left with a potentially solvable Pell equations. We see that $\gcd(y,y-B)$ divides $B$ so we seek solutions $$\{y,y-B\}=\{m\alpha_1u^2,m\alpha_2v^2\}$$ where $m\cdot \gcd(\alpha_1,\alpha_2)$ divides $B$ and $\alpha_1\alpha_2=\alpha.$ 
For example $x,z=3,5$ reduces to having $P(3,y,5)=2\cdot30^2\cdot y\cdot(y-5) $ a square. 
There are no solutions to $\{y-5,y\}=\{u^2,2v^2\}$ (in either order)  but  $\{y-5,y\}=\{5u^2,10v^2\}$ means considering $P(3,5w,5)=2\cdot150^2  \cdot (w-1)\cdot w.$ We need $(w-1)w=2\Box$ with familiar solutions $(w-1)w=1\cdot2,8\cdot 9,49 \cdot 50,288 \cdot 289\cdots.$
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