Solving the equation $f(x+t)=f(x)+f(t)+2\sqrt{f(x)}\sqrt{f(t)}$ I am trying to solve the equation $f(x+t)=f(x)+f(t)+2\sqrt{f(x)}\sqrt{f(t)}$ - as in find a function that satisfies this equation. I notice that the RHS is $({\sqrt{f(x)}+\sqrt{f(t)}})^2$ but I am stuck after this.
 A: Use the hint and solve in the usual step by step way for $g$ first. Assume wlog $g(1)= 1$. 


*

*solve for $x, t$ natural numbers and $0$ (domain can't be
negative) to see $g(n) = n$.

*solve for fractions with numerator being $1$ to see $n\cdot
    g(1/n) = 1$.

*now for rationals to see $q/p \cdot g(p/q) = 1$

*use limit argument for irrationals to get $g(r) = r$

*plug back into $f$ to conclude one solution is $f(n) = n^2$.
A: solution: $f(x) = a^2 x^2$ where $a$ is any real number
A: The presence of $\sqrt{f(x)}$ in your functional equation implies that the range of $f$ is nonnegative.  
If you are looking for a continuous function, Simon's comment shows that $g$ must be linear.  (With real functions, continuity and additivity imply full linearity. This is mentioned here although I'd prefer to show a link to an actual proof.)
Therefore $$\sqrt{f(x)} = ax$$
for some $a$.  


*

*If $a=0$, then $f$ is the zero function.

*If $a>0$, then $f$ is only defined for $x\geq 0$, and $f(x)=a^2x^2$.

*If $a<0$, then $f$ is only defined for $x\leq 0$, and $f(x)=a^2x^2$.


That is, there are three families of solutions.  They all have the same form $f(x)=a^2x^2$, but either you must restrict the domain to non-negatives, non-positives, or $a $ is necessarily $0$.
