$f(a^2)=f(a)^2$ $f(a^2+2ab+b^2)=f(\sqrt{a^2+b^2})^2+f(\sqrt{2ab})^2=f(a^2+b^2)+f(2ab)$ For $x \geq y \geq 0$ we can always find $a,b $ that $x=a^2+b^2,y=2ab,a=\frac{\sqrt{x+y}+\sqrt{x-y}}{2},b=\frac{\sqrt{x+y}-\sqrt{x-y}}{2} $ and so $f(x+y)=f(x)+f(y) \to f(x)=kx$ $f(a)^2=f(a)^2 \to ka^2=k^2a^2 \to f(x)=0,x$

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@above that only covers integers

Let $P(n,x_1,x_2,\ldots,x_n)$ be the given assertion. $P(1,x)\Rightarrow f(x^2)=f(x)^2$ $P\left(2,\sqrt x,\sqrt y\right)\Rightarrow f(x+y)=f(x)+f(y)$, and since $f(x)\ge0$, we will have $f(x)=cx$. Testing, the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=0}$.