Find all solutions to the diophantine equation $(x+2)(y+2)(z+2)=(x+y+z+2)^2$ Solve in postive integer the equation
$$(x+2)(y+2)(z+2)=(x+y+z+2)^2$$
It is rather easy to find several parametric solutions, (such $(a,b,c)=(2,1,1),(2,2,2)$).but it seems harder to find a complete enumeration of all the solutions.
and I have proved it for these some solutions :
since we have $$x^2+y^2+z^2=xyz+4$$
because we have 
$$\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+\left(c+\dfrac{1}{c}\right)^2=\left(a+\dfrac{1}{a}\right)\left(b+\dfrac{1}{b}\right)\left(c+\dfrac{1}{c}\right)+4,abc=1$$
so Let
$$x=a+\dfrac{1}{a},y=b+\dfrac{1}{b},z=c+\dfrac{1}{c},abc=1,a,b,c>0$$
It seems that there are no nontrivial solutions , but I cannot prove this.

For the equation.
$$(x+2)(y+2)(z+2)=(x+y+z+2)^2$$
If we use the solutions of Pell equations.
$$p^2-(z^2-4)s^2=1$$
Using my replacement.
$$a=p^2-2(z+2)ps+(z^2-4)s^2$$
$$b=p^2-2zps+(z^2-4)s^2$$
Decisions will be.
$$x=-2a^2+2(z+2)ab-z(z+2)b^2$$
$$y=-za^2+2(z+2)ab-2(z+2)b^2$$
