Solve the functional equation, $f(x+y+1)= \left(\sqrt{f(x)} + \sqrt{f(y)}\right)^2$ Solve the functional equation, find all real valued functions $f(x)$ s.t.
$$f(x+y+1)= \left(\sqrt{f(x)} + \sqrt{f(y)}\right)^2$$
given $f(0)=1$.
I reached to a point that $4f(x)=f(2x+1)$
 A: Assume that $f$ is a solution and define $g$ by $g(t)=\sqrt{f(t-1)}$, then $g(1)=1$ and $g(t+s)=g(t)+g(s)$ for every $(t,s)$, that is, $g$ is additive.
Linear functions are additive but whether or not adding the hypothesis of continuity is necessary for additive functions to be linear depends on the axiom of choice. That is, with the axiom of choice, there exists additive functions besides the linear ones. For an example, complete $\{1\}$ to a Hamel basis $B$ for $\mathbb{R}$ as a $\mathbb{Q}$ vector space and let $g:\mathbb{R}\to\mathbb{R}$ be the unique linear function such that $g(1)=1$ and $g(b)=0$ for every $b$ in $B$, $b\ne 1$. Every non linear additive function is discontinuous everywhere.
See here or even there for less sketchy explanations.
Edit: Actually, the conditions that $g$ is additive and nonnegative everywhere suffice to conclude. To wit, for every solution $g$, $g(1+0)=g(1)+g(0)$ hence $g(0)=0$, and $g(1-1)=g(1)+g(-1)$ hence $g(-1)=-g(1)=-1$. Since $g(-1)=\sqrt{f(-2)}$, this is absurd, hence the original equation has no solution.
