Following problem though not a research problem


if $x,y,z,w$ are postive integers,and such
$$xyzw=504(x^2+y^2+z^2+w^2)$$


such example $(x,y,z,w)=(21,63,84,84)$ hold,


Now My problem there exist distinct postive integer solution? or find this equation all solution?



Find a distinct postive integer solution to this $xyzw=504(x^2+y^2+z^2+w^2)$ diophantine equation
The basic idea for Markoff-Hurwitz type equations
$$ x\_1^2+\cdots+x\_n^2=Ax\_1x\_2\cdots x\_n $$
with $A\in\mathbb Z$ is that if you have a solution $(x\_1,\ldots,x\_n)$ in integers, then by fixing $n-1$ of the variables to be the given values, you get a monic quadratic equation for the last variable, so since there is one integer solution to that quadratic, there are two. Similarly, if $A$ is a rational number, as in your case, any solution in rational solutions $(x\_1,\ldots,x\_n)$ leads to $n$ new solutions in rational numbers by "flipping" one of the coordinates. (Of course, this is a lie, since various of the $x\_1$ might be the same and/or you might get a quadratic with a single root. But roughly speaking, you tend to get $n$ new solutions.)


You're looking at $x^2+y^2+z^2+w^2=\frac{1}{504}xyzw$. Taking the solution that you give, $(x,y,z,w)=(21,63,84,84)$, we get the flipped solutions
$$ (861,63,84,84),\quad (21,231,84,84),\quad \left(21,63,\frac{273}{2},84\right). $$
So only the first two have integer coordinates. Further flips yield quite large solutions such as
$$
(861,11991,84,84),\quad (3213,231,84,84).
$$
And indeed, it's easy enough to see that if $(x,y,84,84)$ is a solution in integers, then flipping the $x$ or the $y$ gives integer solutions, since setting $z=w=84$ in your original equation yields $x^2+y^2-14xy+14112$, the flip equations for the $x$ and $y$ coordinates are monic. 


