Find all functions satisfying $f(x)f(y)=f(x+y)+xy$

Question

Find all functions satisfying $f(x)f(y)=f(x+y)+xy$.

I know $f(0)=1$ and $f(x)=1\pm x$ are solutions but I'm not sure how to show there aren't any others

Answer 1

We remark that the function which is identically $0$ is not a solution, so we may assume that there is some value $x_0$ for which $f(x_0)\neq 0$.

Taking $x=0,\,y=x_0$ we get $$f(0)f(x_0)=f(x_0)\implies f(0)=1$$

Now letting $y=-x$ we get $$f(x)f(-x)=1-x^2\implies f(1)f(-1)=0$$

Thus at least one of $f(1),\,f(-1)$ is $0$.

Case I: $f(1)=0$.

Then letting $y=1$ we get $$0=f(x+1)+x\implies f(x)=1-x$$

Case II: $f(-1)=0$

letting $y=-1$ we get $$0=f(x-1)-x\implies f(x)=1+x$$