Integeral solutions other than zeros

Relevant wiki: Quadratic Diophantine Equations - Fermat's Method of Infinite Descent

If $a,b,c$ are integer solutions of the equation $a^2 + b^2 + c^2 \; = \; 2^n abc$ where $n \in \mathbb{N}$ , then $a^2 + b^2 +c^2$ is even, so we cannot have all three of $a,b,c$ odd. Thus at least one of $a,b,c$ is even, which means that $a^2 + b^2 + c^2 = 2^n abc$ is a multiple of $4$ . Thus we cannot have one of $a,b,c$ even and the other two odd (since then $a^2 + b^2 + c^2 \equiv 2 \pmod{4}$ ), so it follows that all three of $a,b,c$ are even. Thus we can write $a = 2\alpha$ , $b = 2\beta$ , $c = 2\gamma$ , where $\alpha^2 + \beta^2 + \gamma^2 \; = \; 2^{n+1}\alpha\beta\gamma$ But the same argument will tell us that each of $\alpha,\beta,\gamma$ are even, and we can write $\alpha = 2A,\beta =2B,\gamma = 2C$ where $A^2 + B^2 + C^2 \; = \; 2^{n+2}A B C$ and now $A,B,C$ will all be even, and so on. Being more careful, we could show by induction on $m$ that if $a,b,c \in \mathbb{Z}$ are such that $a^2 + b^2 + c^2 \; = \; 2abc$ then, for any $m \ge 1$ , it is possible to write $a \; = \; 2^m u \;,\; b \; = \; 2^mv \; , \; c \; = \; 2^m w \hspace{1cm} u,v,w \in \mathbb{Z}$ such that $u^2 + v^2 + w^2 \; = \; 2^{m+1}uvw$

Now we are in a position to use a descente infinie argument. If $a \in \mathbb{N}$ , then $a$ can only be divided by $2$ a finite number of times, which contradicts the above.

Thus the only integer solution to the equation $a^2 + b^2 + c^2 \; = \; 2abc$ is $a=b=c=0$ , making the answer to the question $\boxed{0}$ .