$f(x+y+1)=\biggr(\sqrt{f(x)}+\sqrt{f(y)}\biggr)^2$, find $f(x)$. Here is a simple yet a bit queer question.
Let $$f(x+y+1)=\biggr(\sqrt{f(x)}+\sqrt{f(y)}\biggr)^2,$$
and $f(0)=1~~\forall ~x,y \in \Bbb R$.
Find $f(x)$.
I did some work on it and found $f(-1)=0$. A bit more grunt work and I also found the answer to be $(x+1)^2$ and checked it by plugging some numbers.
But I don't know how to solve it properly. I welcome some hints on how to proceed further.
 A: Actually, the function $f_0:\mathbb{R}\to\mathbb{R}_{\geq 0}$ defined by $f_0(x)=(x+1)^2$ isn't a solution to this functional equation, at least not if the domain is $\mathbb{R}$. The reason is that in general $\sqrt{f_0(x)}=|x+1|$, and the absolute value becomes problematic. Indeed, the equation $f_0(x+y+1)=\left(\sqrt{f_0(x)}+\sqrt{f_0(y)}\right)^2$ is, after simplifying, equivalent to the equation
$$
(x+1)(y+1)=|x+1||y+1|,
$$
which fails exactly if $x<-1<y$ or $y<-1<x$. On the other hand, if we restrict $f_0$ to $[-1,\infty[$ or $]-\infty,-1]$, then the functional equation is satisfied.
The original problem actually admits no solution. Indeed, if we plug in $x=y=-1$, then we obtain $f(-1)=4f(-1)$, and hence $f(-1)=0$. But then if we plug in $x=0$ and $y=-2$, we have
$$
0=f(-1)=\left(\sqrt{f(0)}+\sqrt{f(-2)}\right)^{2}=\left(1+\sqrt{f(-2)}\right)^{2}.
$$
This implies $\sqrt{f(-2)}=-1$, which is impossible.
On the other hand, if we are interested in functions $f:I\to\mathbb{R}_{\geq 0}$, where $I=[-1,\infty[$, which satisfy $f(x+y+1)=\left(\sqrt{f(x)}+\sqrt{f(y)}\right)^2$ for all $x,y\in I$ as well as $f(0)=1$, then this admits a unique solution. Indeed, the function $g:\mathbb{R}_{\geq 0}\to \mathbb{R}_{\geq 0}$ defined by $g(x)=\sqrt{f(x-1)}$ is additive. Now if we define $G:\mathbb{R}\to\mathbb{R}$ by the formula
$$
G(x)=
\begin{cases}
 g(x) &\text{if }x\geq 0\\
-g(-x) &\text{ else,}
\end{cases}
$$
then it is straightforward to prove that also $G$ is additive. Furthermore, $G$ is non-negative on $\mathbb{R}_{\geq 0}$, and hence the graph of $G$ isn't dense. From the general theory around the Cauchy-equations, it follows that $G$ must be linear, i.e. of the form $G(x)=cx$ for all $x$, where $c$ is some constant. As $G(1)=1$, we obtain $c=1$, so $G(x)=x$ for all $x$. This then shows that $f(x)=(x+1)^2$ for all $x\in I$, and plugging this into the original equation shows that this is indeed a solution.
A: Let $g(x)=\sqrt{f(x-1)}\ge0$. Then, the function $g(x)$ satisfies the equality $g(x+y)=g(x)+g(y)$, with $g(1)=1$. This is exactly the famous Cauchy functional equation.
