find all functions such that $f(x)\geq0$ for all $x$ and $f(x+t) = f(x) +f(t) +2\sqrt{f(x)f(t)}$

Question

I have thought of $f(x)=x^2$, but it’s not always positive and it doesn’t work when $x$ or $t$ is negative.

Answer 1

We have $$f(x+t)=\left[\sqrt {f(x)}+\sqrt {f(t)}\right]^2$$which leads to$$\sqrt {f(x+t)}=\sqrt {f(x)}+\sqrt {f(t)}$$since $f(x)\ge 0$. By defining $g(x)=\sqrt {f(x)}\ge0$ we obtain $$g(x+t)=g(x)+g(t)$$Therefore the final answer becomes:

$f(x)=g^2(x)$ is the general solution, provided $g(x)$ is a non-negative additive function.

Answer 2

The equation at $x=t=0$ gives us: $f (0)=2*f (0) + 2×f (0) $. So $0 <f (0)=0$. Thus, there isn't any $f $ satisfying this.