Three variable one equation

Clearly $x=y=z=0$ is a solution to the system

Now consider $x\neq0$ $y\neq0$ $z\neq0$

As the right hand side of the equation is even then the left side must also be even. Then either two or none of the variables are odd. First suppose that two of them are odd:

wlog assume $x=2k$ , $y=2n+1$ , $z=2m+1$

Then we have $4k^2+4n^2+4n+1+4m^2+4m+1=4k(2n+1)(2m+1) \implies 4(k^2+n^2+n+m^2+m)+2=4k(2n+1)(2m+1)$ but then the right hand side is a multiple of 4 while the left hand side isn't so we have no solution in this case. Now suppose that all the variables are even:

wlog assume $x=2k$ , $y=2n$ , $z=2m$

Then we have $4k^2+4n^2+4m^2=16knm \implies k^2+n^2+m^2=4knm$ but then this is the same case we had at the beginning (with $k$ , $n$ and $m$ instead of $x$ , $y$ , $z$ ) which means that $k$ , $n$ and $m$ are all even and if we set $k=2k'$ , $n=2n'$ , $m=2m'$ we will get the same case again implying that $k'$ , $n'$ and $m'$ are all even and so on. Thus we get that $x$ , $y$ , $z$ are all powers of 2.

Now let $x=2^k$ , $y=2^n$ , $z=2^m$

Then we have $2^{2k}+2^{2n}+2^{2m}=2^{k+n+m+1}$ so we have that 3 powers of 2 add up to another power of 2. But this can only happen if the powers are of the form $2^a$ , $2^a$ and $2^{a+1}$ ( $2^a+2^a+2^{a+1}=2^{a+2}$ )

But then we have that $(x^2 , y^2 , z^2)=(2^a , 2^a , 2^{a+1})$ and either $2^a$ or $2^{a+1}$ will not have an integral square root.

So we have no solutions and thus the only solution to the equation is $x=y=z=0$

To study the behavior of this equation we must first notice that it is a part of the wider family $x^2 + y^2 + z^2 = 2^w xyz$ , with $w>0$ . Let $S$ be the set of all non-zero solutions of this equation and let's define a function $f: S \to \mathbb{N}$ as follows: $f(x,y,z,w) = |xyzw|$ . Now, let $S' = \{ f(x,y,z,w), (x,y,z,w) \in S \}$

Let's suppose that $x^2 + y^2 + z^2 = 2^w xyz$ is the smallest solution in the sense that $f(x,y,z,w)$ is the smallest element of $S'$ . First of all, as the right-hand side is even, and 3 odd numbers added together would be odd, at least one of the variables is even. By symmetry, and without loss of generality, let $x = 2x_0$ . Then $4x_0^2 + y^2 + z^2 = 2^{w+1}x_0yz$ . Thus, $y^2 + z^2 \equiv 0 \mod 4$ . This is a finite congruence that can be easily solved by checking. The only solutions are $y^2 \equiv z^2 \equiv 0 \mod 4 \implies y,z \equiv 2,4 \mod 4$ . In each case, we have that both variables are also even so let $y = 2y_0, z = 2z_0$ .

Plugging these substitutions into the equation, and dividing by four we get $x_0^2 + y_0^2 + z_0^2 = 2^{w+1}x_0y_0z_0$ . So we have found that if $(x,y,z,w)$ is a solution of the equation, $\left( \frac{x}{2},\frac{y}{2},\frac{z}{2},w+1 \right)$ is also a solution. But $f\left( \frac{x}{2},\frac{y}{2},\frac{z}{2},w+1 \right) = \left| \frac{xyz(w+1)}{8} \right| < |xyzw| = f(x,y,z,w)$ which contradicts that this was the minimal element of $S'$ . Thus, no solution of this equation exists.

However, we assumed at first that $S$ only contained non-zero solutions. It is a trivial exercise to show that if at least one of the variables is 0, all of them are. And that indeed $(0,0,0,1)$ is a solution.

Consider the quadratic in x and find the discriminant. It is clearly less than 0 except when z=y=0, in which case x=0.