That's it?

$\begin{aligned}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{aligned}$ Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm)

$f(x) = f(352 + x)$ So we need only to consider one period $f(0), f(1), ... f(351)$ , which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.

But we also know that $f(x) = f(46 - x) = f(398 - x)$ , so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated. This gives a total of

$352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }$ distinct values.

I'll try to give a somewhat general solution here. Let $a<b<c$ be distinct positive integers and $f$ be a function defined on integers such that for every integer $x$ , $f(x)=f(a-x)=f(b-x)=f(c-x)$ . We would like to know the cardinal of the image of $f$ .

For every integer $x$ , $f(x)=f(b-x)=f(a-b+x)$ and likewise $f(x)=f(b-c+x)=f(c-a+x)$ . Therefore, we can add or substract $a-b,b-c,c-a$ to $x$ without changing its image through $f$ .

Let $d=\mathrm{gcd}(a-b,b-c,c-a)$ and $e=a\bmod d=b\bmod d=c\bmod d$ . For every relative integer $k$ , it follows that $f(x)=f(kd+x)=f(e+kd-x)$ and $f$ is periodic. Let $f^{-1}\{f(x)\}$ the set of all inverse images of $f(x)$ . $f^{-1}\{f(x)\}$ contains therefore $x,e-x,d+x,e+d-x,\dots$

Assuming the cardinal of the image set of $f$ is maximal, and $e\leq d$ for $x=\left\lfloor(e+1)/2\right\rfloor,\dots,\left\lfloor(e+d)/2\right\rfloor$ we have distinct (disjoint) $f^{-1}\{f(x)\}$ (look at $x,e-x,d+x,e+d-x$ , other values won't interfere). Therefore, in this case, the total number of (disjoint) sets $f^{-1}\{f(x)\}$ is $\boxed{\left\lfloor(e+d)/2\right\rfloor-\left\lfloor(e+1)/2\right\rfloor+1}$ .

Plugging in $a=398,b=2158,c=3214$ yields $d=352,e=46$ and therefore there are $\boxed{177}$ different values for $f(x)$ with $x=0,\dots,999$ ( $1000\geq d$ ).

Not easy to write down formally though.