All-Russian Olympiad question (composite of quadratics) From the $1995$ All-Russian Olympiad, $9^{th}$ Graders, Final Round:

Is it possible for the equation $f(g(h(x)))=0$, where $f, g$ and $h$ are quadratic functions, to have solutions $x=1,2,\dots,8$?

I am going to post my own solution so that people can check if it's valid or the most elegant.
 A: No, it is not possible. Suppose this were the case. Put$$f(x) = \alpha(x - a)(x - b).$$Then the polynomials $g(h(x)) - a$ and $g(h(x)) - b$ have $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ as their roots. However, they only differ in their constant term, so in particular their sum of roots is the same. Hence the sum of roots of each is $18$.
Repeat the process, writing$$g(x) - a = \beta(x - c)(x - d).$$Then $h(x) - c$ and $h(x) - d$ again have the same sum of roots, namely $9$. We must then have$$h(1) = h(8), \quad h(2) = h(7), \quad h(3) = h(6), \quad h(4) = h(5).$$We must have$$h(x) = px^2 - 9px + q$$for some $p$, $q$ with $p \neq 0$.
Now we are reduced to finding $f$, $g$ such that$$f(g(x)) = 0$$has roots$$q - 8p, \quad q - 14p, \quad q - 18p, \quad q - 20p.$$Again, a necessary condition is that we can split the roots into two pairs with equal sum, but clearly we can not do that with these four numbers. We conclude $f$, $g$, $h$ do not exist.
A: Essentially, $h$ has to map $8$ values into $4$, $g$ maps $4$ values into $2$, and $f$ maps $2$ values into the single number $0$
Now I'll try to make $f(g(h(x)))$ symmetric around the $y$-axis for ease of calculation. 
If it's possible to have $f(g(h(x)))=(x-1)(x-2)...(x-8)$ it should also be possible to have 
$f_1(g_1(h_1(x)))=(x-3.5)(x-2.5)(x-1.5)(x-0.5)(x+0.5)(x+1.5)(x+2.5)(x+3.5)$
which is simply $f(g(h(x+4.5)$ but $h(x+4.5)$ is also a quadratic in $x$
And consequently $f_2(g_2(h_2(x)))=f_1(g_1(h_1(x/2)))=\dfrac {1} {2^8} (x-7)(x-5)...(x+5)(x+7)$
Again $h_1(x/2)$ is a quadratic in $x$
If $h_2$ is to map $7,5,...,-5,-7$ to $4$ values, it must be symmetric around the y-axis and must be of the form $x^2+a$
But the resulting $4$ values which $g$ has to map is $1+a, 9+a, 25+a, 49+a$, and to map them into $2$ values, points $1+a$ and $49+a$ have to be equidistant from some $c$, and so does $9+a$ and $25+a$ (but from the same $c$). Clearly there is no such $c$.
