How to solve equation in integer numbers? I have following equation:
$$x^2 + y^2 + z^2 + v^2 = 2 x y z v.$$
How can I solve this one? Roots should be integer numbers.
Edit: I tried to prove, that given equation don't have solutions. I fixed $z, v$ variables. Then task boils down to finding intersection between paraboloid and saddle. Unfortunately, it is hard way. Maybe there are another ways/patterns for solving similar equations?
 A: Outline: It will turn out that the only solution is with all the variables equal to $0$.
The right-hand side is even. So it is clear our variables are not all odd. 
If two are odd and two are even, the left-hand side is congruent to $2$ modulo $4$, but the right-hand side is divisible by $4$, impossible.
So all the variables are even. Let $x=2x_1$, $y=2y_1$, and so on.
Substitute in our equation and simplify. We get $$x_1^2+y_1^2+z_1^2+v_1^2=8x_1y_1z_1v_1.$$
Now repeat the argument. We find that $x_1,y_1,z_1,v_1$ are all even. Let $x_1=2x_2$, $y_1=2y_2$, and so on. We get
$$x_2^2+y_2^2+z_2^2+v_2^2=32x_2y_2z_2v_2.$$
Continue. We conclude that each of $x,y,z,v$ is divisible by arbitrarily high powers of $2$, so all of them are $0$.
Remark: This sort of argument is usually called an infinite descent argument. It is ordinarily credited to Fermat. It is essentially an induction argument. 
If you prefer, you can write it up as follows. If there is a solution with the variables not  all equal to $0$, then there is a solution with $|x|+|y|+|z|+|v|$ positive and minimal. But then $(x_1,y_1,z_1,v_1)$ is a solution with sum of the absolute values greater than $0$ but less than $|x|+|y|+|z|+|v|$: Contradiction!
A: On the other hand, there are infinitely many solutions to 
$$  x^2 + y^2 + z^2 + v^2 = 4 xyzv,$$
occurring in a tree similar to the tree of Markov Numbers. It took me a while to find out, but most Markov things are the same person. 
There are also infinitely many solutions to 
$$  x^2 + y^2 + z^2 + v^2 = 4 xyzv,$$
although this time all the variables are even, dividing all of them by $2$ gives a quadruple in the first tree. 
For details, see  HURWITZ 
