Non-negative integer

We can easily see that $(0,0,0,0)$ is a solution. We'll prove there are no others. The proof uses a popular method called Proof by Infinite Descent .

Proof by contradiction:

Assume it is possible that at least one of $x,y,z,w$ is positive.

The RHS is a perfect square, therefore $\text{RHS}\equiv (xyz)^2\equiv \begin{cases}0 && \text{if }xyz\text{ is even.}\\\text{or}\\1 && \text{if }xyz\text{ is odd.}\end{cases}\pmod 4$

Similarly, the same holds for $x^2,y^2,z^2,w^2$ . So we can only have all of them even, all of them odd or 3 even and 1 odd.

If all of them are odd, then the LHS is even and the RHS is odd, contradiction.

If 3 of them are even and 1 of them is odd, then the LHS is odd and the RHS is even, contradiction.

Therefore, all of them are even. Let $x=2a,y=2b,z=2c,w=2d$ .

$a^2+b^2+c^2+d^2=16a^2b^2c^2$

The RHS is divisible by 4, hence the LHS is divisible by 4. Therefore, all of them are even or all of them are odd. If all are odd, then the LHS is not divisible by 8, while the RHS is.

Hence $a,b,c,d$ are even. Let $a=2k, b=2l, c=2m, d=2n$ .

$k^2+l^2+m^2+n^2=256k^2l^2m^2$

The same argument continues - we can easily see that we can apply the same arguments we did with the last case, since the RHS is still divisible by 4 and 8 and the LHS is still a sum of 4 squares.

This pattern will continue - we will always have all of the variables even (in this case we have $k,l,m,n$ even). The ones that were initially positive (i.e. non-zero) will get continuously reduced, which is impossible (you cannot just divide a positive integer by 2 forever and get an integer each time).

Hence contradiction - there cannot be any positive solutions, since they would get reduced forever, while 0 does not get reduced when divided by 2. $\square$