Equation with three variables I am confused as how to solve an equation with three squared variables to get its integer solutions? As: 
$$x^2+y^2+z^2=200$$
Thanks!
 A: All of $x^2$, $y^2$, and $z^2$ must be non-negative, so there are only finitely many cases to check. 
A: If you are doing it to find solution to your previous question about the area of that triangle then you should use heron's formula and simplify it in terms of $x, y, z$ like,
$\sqrt{s(s-a)(s-b)(s-c)}=\dfrac{1}{4}\sqrt{(x+y+z)(x+y-z)(x-y+z)(-x+y+z)}$, which will simplify into,
$\dfrac{1}{4}\sqrt{-(x^4+y^4+z^4)+2(x^2y^2+y^2z^2+z^2x^2)}$
Now you can substitute the values to get the required area.
A: I would do it this way: (it takes just few minutes)
Let's assume $x<=y<=z$, without loss of generality.
$z$ must be larger than $8$, since $3*8^2 = 192 < 200$.
$z$ must be smaller than $15$, since $15^2 = 225 > 200$.
Now, there are six possibilities:
$$x^2+y^2=200-9^2=119$$
$$x^2+y^2=200-10^2=100$$
$$x^2+y^2=200-11^2=79$$
$$x^2+y^2=200-12^2=56$$
$$x^2+y^2=200-13^2=31$$
$$x^2+y^2=200-14^2=4$$
Then, according to the list of Pythagorean triples, only this solution can exist:

$$x=6, y=8, z=10$$

The list above assumes x<>0, and for x=0, we easily get two more solutions:

$$x=0, y=2, z=14$$

and

$$x=0, y=10, z=10$$

Of course, if variables can be negative, there is also a number of solutions directly derived from last three ones.
