Team Selection Test for IMO 2019

Let $P(x)=A(x-r_1{)}^{d_1}\cdots (x-r_k{)}^{d_k}$ where $r_1<\cdots <r_k$. It is easy to see that $Q(x)$ has degree $2$ and hence $Q(x)=a{x}^2+bx+c$ for some real numbers $a$, $b$ and $c$. Then the given equality can be written as $$A^2\prod _{i=1}^k(x-r_i{)}^{2d_i}=A\prod _{i=1}^k(a{x}^2+bx+c-r_i{)}^{d_i}.$$ Therefore, for each $i$ the roots of $a{x}^2+bx+c-r_i$ are $r_s$ and $r_t$ for some $s$ and $t$. On the other hand, the sum of the roots are $-b/a$ for every $i$. Thus, all the roots of $(P(x){)}^2$ can be paired in a way that sum of the elements in each pair is the same. Let an $r_1$ be

paired with $r_s$ and an $r_k$ be paired with $r_t$. Since $r_1\le r_t$, $r_s\le r_k$ and $r_1+r_s=r_k+r_t$, we see that $r_s=r_k$ and $r_t=r_1$. In other words, every $r_1$ has to be paired with an $r_k$ and every $r_k$ has to be matched with an $r_1$. Therefore, we obtain that $d_1=d_k$. By induction it is easy to prove that $d_i=d_{k+1-i}$ for every $i$ and all pairs are of the form $\{r_j,r_{k+1-j}\}$. Consequently, for every $m$ we have that $(c-r_m)/a$ is equal to $r_j{r}_{k+1-j}$ for some $j$. However, the numbers of the form $r_j{r}_{k+1-j}$ can attain at most $\lfloor \frac{k+1}{2}\rfloor$ distinct values and hence we get $k\le \lfloor \frac{k+1}{2}\rfloor$ which implies $k\le 1$. Therefore, all the roots of $P(x)$ are the same.