Functional equation defined over non-negative real numbers: $f^{-1}\big(f(x)f(y)\big)=\frac{f(x+y)-f(x)-f(y)}2$ I'm new to this forum and I don't know how to write mathematical symbols.
I have the following functional equation:

*

*$f$ defined on $[0, +\infty)$ with values in $[0, +\infty)$

*$f$ is
bijective and the following relation holds (*):

$$f^{-1}\big(f(x)f(y)\big)=\frac{f(x+y)-f(x)-f(y)}2$$ Find all functions with this property. I found that $f(0)=0$ with standard manipulations. $f$ is also continuous (comments). Thank you!
 A: It seems the following.  
I shall continue from already obtained results. Since the function $f$ is bijective, there exists a point $x_0$ such that $f(x_0)=1$. Putting in the relation (*) $x=x_0$ and $y=nx_0$ we obtain
$$f^{-1}(f(x_0)f(nx_0))=\frac{f((n+1)x_0)-f(x_0)-f(nx_0)}2$$
$$nx_0=\frac{f((n+1)x_0)-1-f(nx_0)}2$$
$$f((n+1)x_0)= f(nx_0)+2nx_0+1$$
From this recurrence, using initial condition $f(x_0)=1$ we obtain
$$f(nx_0)=n+n(n-1)x_0.$$
Putting in the relation (*) $x=2x_0$ and $y=2x_0$ we obtain
$$f^{-1}(f(2x_0)^2)=\frac{f(4x_0)-2f(2x_0)}2$$
$$f^{-1}((2+2x_0)^2)=\frac{4+12x_0-2(2+2x_0)}2$$
$$f^{-1}((2+2x_0)^2)=4x_0$$
$$(2+2x_0)^2=f(4x_0)=4+12x_0$$
$$(1+x_0)^2=1+3x_0$$
$$x_0^2=x_0$$
$$x_0=1$$
So 
$$f(n)=n^2.$$
Putting in the relation (*) $x=1$ and $y=y$ we obtain
$$f^{-1}(f(y))=\frac{f(y+1)-f(1)-f(y)}2$$
$$2y=f(y+1)-1-f(y)$$
$$f(y+1)=f(y)+2y+1$$
Iterating this recurrence, we obtain (**)
$$f(y+n)=f(y)+2ny+n^2.$$
Putting in the relation (*) $x=n$ and $y=y$ we obtain
$$f^{-1}(f(n)f(y))=\frac{f(n+y)-f(n)-f(y)}2$$
$$f^{-1}(n^2f(y))=\frac{f(n+y)-n^2-f(y)}2$$
Using (**), we obtain 
$$f^{-1}(n^2f(y))=ny$$
$$n^2f(y)=f(ny).$$
Therefore for each positive rational number $\frac pq$ we have 
$$f\left(\frac pq\right)=\frac 1{q^2}f(p)= \frac {p^2}{q^2}.$$
Since the function $f$ is continuous, for each $x\ge 0$ we have 
$$f(x)=x^2.$$
