Find $n \in N$, for which $a+b+c+d = n \sqrt{(abcd)}$ has a solution $(a, b, c, d)$ in natural numbers. By pure logical observations, I got that $n=1,2,4$ satisfies but I couldn't get the proper theoretical proof. Any help is appreciated. 
 A: I generalized to 
$$  (w+x+y+z)^2 = k wxyz   $$
A solution is "fundamental" in the sense of Hurwitz if  we demand 
$$ w \geq x \geq y \geq z \geq 1  $$
and
$$  kxyz \geq 2(w+x+y+z) \; . \;   $$
Any solution for a certain value of $k$ descends to a fundamental by "Vieta Jumping." Thus, if there are no fundamental solutions for a given $k,$ there are no solutions at all.
ADDED: just as with the tree of Markov Numbers, any of the fundamental solutions indicated below becomes the root of a tree of solutions. We move around within the tree by Vieta jumping, going in different directions depending on which number we choose to do the jumping. If we are jumping the entry called $w,$ the jumped value is
$$ w' = kxyz - 2x-2y-2z-w  $$
For example, with my $k=9,$ Batominovski's $(25,2,2,1)$
descends to
$$ (1,2,2,1) $$
since $9 \cdot 2 \cdot 2 -4-4-2 -25 = 36-10-25=1$
These are all I get:
jagy@phobeusjunior:~$ ./mse
  k  1         w 4 x 4 y 4 z 4
  k  1         w 6 x 6 y 3 z 3
  k  1         w 8 x 5 y 5 z 2
  k  1         w 10 x 10 y 9 z 1
  k  1         w 12 x 6 y 4 z 2
  k  1         w 15 x 10 y 3 z 2
  k  1         w 18 x 9 y 8 z 1
  k  1         w 21 x 14 y 6 z 1
  k  1         w 30 x 24 y 5 z 1
  k  2         w 4 x 3 y 3 z 2
  k  2         w 8 x 4 y 2 z 2
  k  2         w 10 x 5 y 4 z 1
  k  2         w 12 x 8 y 3 z 1
  k  3         w 4 x 4 y 3 z 1
  k  3         w 6 x 2 y 2 z 2
  k  3         w 9 x 6 y 2 z 1
  k  4         w 2 x 2 y 2 z 2
  k  4         w 6 x 3 y 2 z 1
  k  5         w 5 x 2 y 2 z 1
  k  5         w 10 x 8 y 1 z 1
  k  6         w 6 x 4 y 1 z 1
  k  8         w 4 x 2 y 1 z 1
  k  9         w 2 x 2 y 1 z 1
  k  12         w 3 x 1 y 1 z 1
  k  16         w 1 x 1 y 1 z 1
Mon Oct 14 12:40:44 PDT 2019
jagy@phobeusjunior:~$ 

