Diophantine quartic equation in four variables
Solution 1:
The equation,
$$wxyz=(w+x+y+z)^2\tag{1}$$
can be solved as a quadratic in $z$,
$$(w + x + y)^2 + \big(2(w + x + y) - w x y\big) z + z^2=0\tag{2}$$
Its discriminant is $wxy(wxy-4\big(w+x+y)\big)$ and must be made a square. It can be shown an initial solution $a,x,y$ can generate an infinite more. Define,
$$axy(axy-4\big(a+x+y)\big)=c^2\tag{3}$$
$$a^2xy(xy-4) =d\tag{4}$$
then a family of solutions to (2) can be given as,
$$w =\frac{a(p-cq)^2}{p^2-dq^2}\tag{5}$$
$$z =-(w+x+y)+\frac{1}{2}\left(wxy\pm\frac{(p-cq)(cp-dq)}{p^2-dq^2}\right)\tag{6}$$
for arbitrary $p,q$. If we want integers, then one can solve the Pell equation $p^2-dq^2=1$. Some examples are,
$$w,x,y = 4,\;4,\;4(p-q)^2;\;\;z=4(p+q)^2\;\text{or}\;\;4(3p-5q)^2,\;\;\text{where}\;p^2-3q^2=1\tag{7}$$
or,
$$w,x,y = 3,\;3,\;6(p-q)^2;\;\;z=6(p+q)^2\;\text{or}\;\;24(p-2q)^2,\;\;\text{where}\;p^2-5q^2=1\tag{8}$$
and so on.
Solution 2:
Your problem meets the requirements that cause the Markov numbers to occur in a tree. Your equation is $$ w^2 + x^2 + y^2 + z^2 = -2(wx+wy+wz+xy+xz+yz) + wxyz. $$ The critical thing is that each variable occurs only to the first power on the right hand side. That is, each term on the right hand side is squarefree as far as the variables are concerned. What happens next is probably known to most users of this site as Vieta Jumping.
Suppose i have a solution with all integer values. I would like to keep $x,y,z$ the same and "jump" $w.$ Well, $$ w^2 + (2x+2y+2z - xyz ) w + (x+y+z)^2 = 0. $$ The sum of the two solutions to this equation in $w$ is $xyz - 2x-2y-2z.$ So we get a new value for $w,$ call it $$ w' = xyz - 2x-2y-2z - w. $$ That is, given a solution $(w,x,y,z),$ we get a new solution with $$ (xyz - 2x-2y-2z - w, x,y,z). $$ If we jump $w$ again, we get back to where we were. The other jumps are $$ (w,wyz - 2w-2y-2z - x, y,z), $$ $$ (w,x, wxz - 2w-2x-2z - y, z), $$ $$ (w,x,y, wxy - 2w-2x-2y - z). $$
What this means is that any solution is connected to an infinite number of solutions. After that things become difficult. In the generalization by Hurwitz, the solutions (for any fixed dimension) are arranged into one or more trees. For dimension three, Markov, there is just the one tree. It requires proof to be sure that the entire set of solutions splits into a forest. For aspects of Markov-Hurwitz, see https://mathoverflow.net/questions/84927/conjecture-on-markov-hurwitz-diophantine-equation
This is kind of interesting. I will try to see what can be done in the way of "fundamental solutions" as in the methods of Hurwitz.
Here are some solutions with positive but small entries, required in descending order.. It is possible to have the gcd of the four variables to be one. Note that jumping the first position in $(4,4,4,4)$ is $(36,4,4,4).$ Jumping the first position in $(10,10,9,1)$ is $(40,10,9,1).$ Jumping the third position in $(10,10,9,1)$ is $(10,10,49,1),$ which can then be ordered into $(49,10,10,1)$ if desired.
Alright, did over allowing negative entries, but still descending in absolute value. It would appear that these, and solutions with one or more entries equal to zero, need to be considered together to get anything sensible. This can be done, though, and was done for the Apollonian Gasket problem. It turns out I forgot to put some absolute values in the program. Better below. I ruled out solutions with any entry $0,$ far too numerous.
2 2 -1 -1
4 4 4 4
6 3 -2 -1
6 6 3 3
8 5 5 2
9 -2 -2 1
10 10 9 1
12 4 -3 -1
12 6 4 2
15 10 3 2
16 4 -2 -2
16 12 -3 -1
18 5 5 2
18 9 8 1
20 5 -4 -1
21 14 6 1
24 6 3 3
25 9 -2 -2
30 5 -3 -2
30 6 -5 -1
30 24 5 1
32 -9 2 -1
32 -6 -3 1
36 4 4 4
36 16 -2 -2
40 10 9 1
40 15 3 2
42 7 -6 -1
45 8 5 2
45 20 -4 -1
48 6 -4 -2
49 -6 -3 2
49 10 10 1
49 25 -2 -2
54 6 -3 -3
54 12 4 2
54 30 5 1
56 8 -7 -1
56 21 6 1
64 36 -2 -2
70 7 -5 -2
72 9 -8 -1
75 -12 -4 1
75 6 6 3
80 45 -4 -1
81 18 8 1
81 49 -2 -2
84 7 -4 -3
90 10 -9 -1
96 8 -6 -2
96 30 -5 -1
98 18 9 1
100 12 6 2
100 64 -2 -2
Solution 3:
Got it, simple observations. If there are no zero entries, there must be an even number of negative entries. Alright, if $(w,x,y,z)$ is a solution, so is $(-w,-x,-y,-z).$ If we have a solution with positive entries, a jump cannot result in zero, as the sum of the entries is still positive. A jump cannot result in a negative entry, as that would be an odd number (one) of negative entries. So, in fact, it is reasonable to consider positive entries.
A Hurwitz fundamental solution is then one for which jumping any entry increases that entry. As we may take the variables in decreasing order, this means that an (ordered) fundamental solution is one with $$ w \geq x \geq y \geq z \geq 1, \; \; \; xyz \geq 2(w+x+y+z). $$
The first few are
w x y z xyz 2(w+x+y+z)
4 4 4 4 64 32
6 6 3 3 52 36
8 5 5 2 50 40
10 10 9 1 90 60
12 6 4 2 48 48 WOW
15 10 3 2 60 60 WOW
18 9 8 1 72 72 WOW
21 14 6 1 84 84 WOW
30 24 5 1 120 120 WOW
where I put WOW to indicate equality. But that just means jumping the largest value keeps it the same, jumping the other variables still changes things. So, the questions become: (A) do the orbits of these solutions under jumping stay distinct, in which case we have a genuine forest? (B) is the number of fundamental solutions finite? (B) is false for Apollonian problem...
EDIIITTTTT: question (B) turns out to be true, and the nine fundamental solutions displayed are all of them. See the answer at Inequality with four positive integers looking for upper bound by user leshik.
See if I can get this across: any positive integer solution can be reduced to a fundamental solution by jumping (the largest current entry) and re-ordering. Don't know yet: can it be reduced to more than one fundamental solution by different choices? This is question (A) above.
Oh, my first impression is that $\gcd(w,x,y,z)$ does not change by jumping, so that separates some trees at any rate. Your first two examples have gcd 4 and 3 respectively.