$f(g(h(x)))=0$ has $8$ real roots

Find all quadratic polynomials $f(x),g(x)$ and $h(x)$ such that the polynomial $f(g(h(x)))=0$ has roots $1,2,3,4,5,6,7$ and $8$.

I don't know what to do. Making a $8$ degree equation is quite tedious. Thanks.

Solution 1:

For any two distinct numbers $r_1, r_2$, there is a quadratic $f$ with these as roots, and in fact the general such $f$ is $a (x - r_1)(x - r_2)$ where $a$ is a nonzero constant. So now you want $g(h(x))$ to take value $r_1$ on four of $1,2,\ldots,8$ and $r_2$ on the other four. Now the four $x$'s (let's say $x_1, \ldots, x_4$) for which $g(h(x)) = r_1$ must correspond to two different values of $h$, say $s_1, s_2$ with $g(s_1) = g(s_2) = r_1$, $h(x_1) = h(x_2) = s_1$ and $h(x_3) = h(x_4) = s_2$; similarly the four $x$'s ($x_5$ to $x_8$) for which $g(h(x)) = r_2$ must correspond to two other values of $h$, with $h(x_5) = h(x_6) = s_3$, $h(x_7) = h(x_8) = s_4$, $g(s_3) = g(s_4) =r_2$. Now two points where a quadratic takes the same value must be symmetric with respect to the critical point of the quadratic. Thus $x_1 + x_2 = x_3 + x_4 = x_5 + x_6 = x_7 + x_8$.

But $x_1, x_2, \ldots, x_8$ must be $1,2,\ldots, 8$ in some order. The only way to split up $1,2,\ldots,8$ into pairs with the same sum is $(1,8), (2,7), (3,6), (4,5)$. So let's say $x_1 = 1$ and $x_2 = 8$. Thus $h(x) = a (x - 9/2)^2$ for some constant $a$, making $h(1)=h(8) = 49 a/4$, $h(2) = h(7) = 25a/4$, $h(3) = h(6) = 9a/4$, $h(4)=h(5) = a/4$. But those four values can't be split into two pairs with the same sum. So the problem (assuming you're asking for non-constant quadratics) has no solution. Of course there are lots of solutions where $f$, $g$ or $h$ is constant.

Are you sure you don't want the roots to be, say, $1,2,5,8,10,13,16,17$?

Solution 2:

The polynomial $$F(x)=\prod_{1\le k\le 8}(x-k)$$ has a derivative equal to $$4(2x-9)(x^6-27x^5+288x^4-1539x^3+4299x^2-5886x+3044)$$ where the second 6-degree factor is irreducible (verify this, for example, in WolframAlpha).

Therefore if $F(x)=f(g(h(x)))$ then $$F’(x)=f’(g(h(x)))\cdot g’(h(x))\cdot h’(x)$$ which is a contradiction because $F’(x)$ is a product of only two irreducible factors.