Let $x_1\leq x_2\leq x_3\leq x_4$ be real numbers. Prove that there exist polynomials of degree two $P(x)$ and $Q(x)$ with real coefficients such that $x_1$, $x_2$, $x_3$ and $x_4$ are the roots of $P(Q(x))$ if and only if $x_1+x_4=x_2+x_3$.
Problem
Source: Spain MO 2023 P4
Tags: algebra, polynomial, Spain
Assassino9931
12.03.2023 11:27
If $a_1, a_2$ are the roots of $P$, then the $x$-s are the roots of $Q(x) - a_i$ and the conditions follows from Vieta's sum of roots formula. For construction take $Q(x) = x^2 - sx + p$ and $a_1 = 0$, $a_2 = p-q$ where $s$ is the common sum, $p = x_1x_2$ and $q = x_3x_4$.
Alfombra12
16.07.2023 23:41
I claim that, for $x_1+x_4=a=x_2+x_3$ and any real $b$, $Q(x)=x^2-ax+b$ and $P(x)=(x-Q(x_1))(x-Q(x_2))$ satisfy the condition.
($\implies$ ) Note that if $x_1,x_2,x_3,x_4$ are the roots of $P(Q(x))$ then $Q(x_1),Q(x_2),Q(x_3),Q(x_4)$ are the roots of $P(x)$; however, given that $P(x)$ only has two roots and the symmetry a quadratic equation has, this implies $Q(x_1)=Q(x_4)$ and $Q(x_2)=Q(x_3)$. Now, let $a$ and $b$ be any real numbers and $Q(x)=x^2-ax+b$, given the symmetry of parabolas we get $x_1+x_4=a=x_2+x_3$.
( $\Longleftarrow$ ) Let $x_1+x_4=a=x_2+x_3$ and define again $Q(x)=x^2-ax+b$ for any real $b$. Note $Q(x_1)=Q(x_4)$ and $Q(x_2)=Q(x_3)$ so let $P(x)=(x-Q(x_1))(x-Q(x_2))$ and thus $x_1,x_2,x_3,x_4$ are the roots of $P(Q(x))$.