Which values of $f$ satisfying $f(x)=xf(x^2-3)-x$ we know? My question is inspired by Functional puzzle: find $f(2)$
A function $f\colon \mathbb{R}\to\mathbb{R}$ satisfies
$$f(x)=xf(x^2-3)-x$$
for all $x$. For which $x$  is the value of $f(x)$ known?
I know values of $f(x)$ for $x\in\{0,\pm1,\pm2\}$. 
Edit: As @StevenStadnicki observed, a domain may be smaller than reals, so every suggestion how large it can be is also interesting.
 A: This question is actually less well-formed than you think it is.  First of all, there's the question of what 'for all $x$ means', because that isn't actually enough to specify a domain. For all $x\in\mathbb{Z}$? For all $x$ in $\mathbb{R}$? A function isn't a function unless you specify the domain and the range, and this question actually does neither.
But beyond that, you say a function, and that's actually another issue; we don't know if any such $f$ exists, and we don't know whether just one $f$ exists.
In the case where the domain of $f()$ is $\mathbb{Z}$, in particular, it seems clear that there are many such functions: since $x^2-3$ as a function is one-to-one from $n\gt 2$ to (but not onto) $\mathbb{N}$, any 'forced' value of $f()$ in this region is forced by only one previous value, and there are no further dependencies; $f(6)$ is defined by $f(3)$, $f(13)$ is defined by $f(4)$, etc. Any value of $f()$ that's not forced by virtue of having an argument of the form $n^2-3$ is completely free and can be taken to be anything whatsoever.  Since 'most' (infinitely many) integers are not of the form $n^2-3$, this gives infinite degrees of freedom; there are either $2^\omega$ or $2^\mathfrak{c}$ such functions, depending on whether your range is $\mathbb{Q}$ or $\mathbb{R}$.
The case where the domain of $f()$ is $\mathbb{R}$ is rather more interesting; now $f()$ is one-to-one for $x\gt x_0$, where $x_0$ is the (largest) solution of $x_0^2-3 = x_0$; this turns out (by a quick quadratic theorem application) to be $x_0=\frac12(1+\sqrt{13})\approx 2.3$. So in principle (see below!) we can pick any (half-open) interval $[x, x^2-3)$ for $x\gt x_0$, choose an arbitrary $f()$ on this range, and then 'replicate' it up and down using the functional equation to get values of $f(x)$ for $x\in(x_0,\infty)$.  You can even get continuity on this domain by requiring that $\lim_{t\to (x^2-3)^-}f(t)$ satisfy the functional equation.  The oddness of $f()$ can then be used to reflect it over to $x\in(-\infty, -x_0)$.
The catch with the above argument is that the 'excitement' in the middle can actually ripple outwards; in particular, for values of $x$ very near zero, we get values of $x^2-3$ near $-3$ and thus get a functional equation for $f()$ near $x=-3$ - or again, by reflection, near $x=3$. So not every arbitrary function will 'reflect down' to a valid one. Instead, it would take a careful analysis of the behavior of intervals under the mapping $x\mapsto x^2-3$ (and even moreso, the behavior under the inverse mapping) to determine 'whether 'fundamental regions' exist which can be replicated around. I suspect the structure would turn out to be somewhat Cantor-like, but it's hard for me to directly 'see' the relevant intervals at the moment.
A: If you know $x=0$ then you know $x^2-3=0\implies x=\pm\sqrt 3$ because:
$f(\pm \sqrt 3)=\pm \sqrt 3f(0)-\pm \sqrt 3$
Now you can solve for $x^2-3=\pm\sqrt3$ with the same logic, and so on.
With this you can create few sets of answers, with those you may can work to more sets(I didn't try to find the values for those sets so idk if you can do further calculations)
