Functional Equation involving Fractional Part: $f(\{a+b\})=\{f(a)f(b)\}$ I'm stuck on the following problem:

Find (or classify) all functions $f$ with the property
  $$f(\{a+b\})=\{f(a)f(b)\}$$
  Where $\{\}$ is the "fractional part" function.

So far, I've determined that 
$$f(0)=0$$ 
and that 
$$f(a)f(-a) \in \mathbb Z, \forall a \in \mathbb R$$
Can anyone help?
If it helps, I solved a similar functional equation
$$f(a+\{b\})=f(b+\{a\})$$
and found that the solution set of functions was the set of functions with period $1$, or a period that goes evenly into one, like $0.5$ or $0.2$.
Any help is appreciated!

This is just a partial solution to the question
We can rewrite the functional equation as
$$f(\{x\})=\{f(x-z)f(z)\},\quad \forall \{x\},z,x-z\in\operatorname{dom}(f)\tag1$$
Suppose that $0\in\operatorname{dom}(f)$, then we found that
$$f(0)=\{f(0)^2\}\implies |f(0)|<1\implies f(0)=f(0)^2\implies f(0)=0\tag2$$
By $(1)$ we have that $f(\{x\})=\{f(x)f(0)\}=0$, what imply that $f$ is zero if $|x|<1$. In general we have that
$$f(x)f(y)\in\Bbb Z,\quad \forall \{x+y\},x,y\in\operatorname{dom}(f)$$
There are infinite solutions from the assumption on the domain of $f$. Just we need that


*

*$f(x)=0$ when $|x|<1$ and

*$f(x)\in\Bbb Z$ when $|x|\ge 1$
