How to find the value of this function If $f:\mathbb{R} \rightarrow\mathbb{R}$ is a function which satisfies $$f(x)+f(y)=\frac{f(x-y)}{2}\cdot \frac{f(x+y)}{2}$$ for all $x\in\mathbb{R}$ and $f(1)=3$, what is $f(6)$?
 A: The only function satisfying the given functional equation is the zero function. Hence the condition $f(1)=3$ cannot hold.
For any $a,z$, we find
$$\frac{ f(z)}2\frac{ f(a)}2=f\left(\frac{z+a}2\right)+f\left(\frac{a-z}2\right)=f\left(\frac{a-z}2\right)+f\left(\frac{z+a}2\right)=\frac{ f(-z)}2\frac{ f(a)}2$$
hence if there exists $a$ with $f(a)\ne0$, the function $f$ is odd, and if there is no such $a$, then $f$ is identically zero (and trivially odd).
We conclude that $f(0)=0$.
Then with $y=0$, we get
$$ f(x)=\frac14f(x)^2$$
hence $f(x)\in\{0,4\}$ for all $x$. Since $f$ is odd, there cannot be $x$ with $f(x)=4$ as $f(-x)=-4$ is excluded. Therefore $f(x)=0$ for all $x$.
A: $$f(1)+f(0)=\frac{1}{4}f^2(1)\implies 4f(0)=f^2(1)-4f(1)=9-12=-3\implies f(0)=-\frac{3}{4} $$
$$f(6)+f(0)=\frac{1}{4}f^2(6)\implies f^2(6)-4f(6)-4f(0)=0\implies f^2(6)-4f(6)+3=0$$
and now solve the equation to conclude...
A: The function you described does not exist because no value of $f(0)$ makes sense:


*

*$f(0) + f(0) = \frac{f(0 - 0)}{2} \cdot \frac{f(0+0)}{2} = \frac{f(0)^2}4$
implies that $8f(0) = f(0)^2$ which means $f(0)$ can either be $0$ or $8$.

*$3+f(0) = f(1) + f(0) = \frac{f(1-0)}{2}\cdot \frac{f(1+0)}{2} = \frac{f(1)^2}{4} = \frac{9}{4}$ means that $f(0) = \frac{9}{4} - 3 = -\frac34$, which leads to a contradiction.

