Is it possible for quadratic functions $f, g, h$ that $f(g(h(x))) = 0$ to have roots as $1,2,3,4,5,6,7,8$?

Question

I tried to factorise $(x-1)\cdot(x-2)\cdots(x-8)$ But it did not yield anything. Any hint will be appreciated. (The answer is given to be NO but how to arrive at it?)

Answer 1

Clearly, we want $h(1),h(2),\cdots,h(8)$ to be only four distinct numbers. This forces (up to multiplication by a constant) $h(x)=(x-4.5)^2$.

Now, calculate the four different values of $h(1),h(2),\cdots,h(8)$. Is there any way you can construct $g$ to make those four into two?