Let $Q(x)$ have leading coefficient $c$, $P(x)$ have leading coefficient $t,$ $P$ have degree $d$, and the roots of $P$ be $r_1 \ge r_2 \ge \cdots \ge r_d.$ Then notice that $t \prod (Q(x)-r_i) = t^2 \prod (x - r_i)^2.$ Therefore, we know that $c^d = t.$ Now, notice that each $Q(x) - r_i$ has roots among the $r_i$'s, and so we can write $Q(x) - r_i = c(x-r_{i_1})(x-r_{i_2}),$ for $r_{i_1} \ge r_{i_2}.$ Now, notice that $r_{i_1} + r_{i_2}$ is constant over all $i$, and so hence the roots of $P$ can be paired into groups $(r_{i_1}, r_{i_2})$ such that the sum of each group is the same and $Q(x) - r_i = c(x-r_{i_1})(x-r_{i_2}).$ Notice that the number of distinct pairs is the same as the number of distinct values among the $r_i$'s, since the number of distinct $Q(x) - r_i$'s is the same as the number of distinct $r_i$'s. It's easily seen that this implies that all the roots are equal. $\square$

This problem is also true when $P(x)$ is a polynomial with complex coefficients

Pathological wrote: Let $Q(x)$ have leading coefficient $c$, $P(x)$ have leading coefficient $t,$ $P$ have degree $d$, and the roots of $P$ be $r_1 \ge r_2 \ge \cdots \ge r_d.$ Then notice that $t \prod (Q(x)-r_i) = t^2 \prod (x - r_i)^2.$ Therefore, we know that $c^d = t.$ Now, notice that each $Q(x) - r_i$ has roots among the $r_i$'s, and so we can write $Q(x) - r_i = c(x-r_{i_1})(x-r_{i_2}),$ for $r_{i_1} \ge r_{i_2}.$ Now, notice that $r_{i_1} + r_{i_2}$ is constant over all $i$, and so hence the roots of $P$ can be paired into groups $(r_{i_1}, r_{i_2})$ such that the sum of each group is the same and $Q(x) - r_i = c(x-r_{i_1})(x-r_{i_2}).$ Notice that the number of distinct pairs is the same as the number of distinct values among the $r_i$'s, since the number of distinct $Q(x) - r_i$'s is the same as the number of distinct $r_i$'s. It's easily seen that this implies that all the roots are equal. $\square$ What is $r_i$'s and why $Q(x) - r_i = c(x-r_{i_1})(x-r_{i_2}) ( you mean deg Q=2) ?? please answer my question

@above yup bro, $degQ=2$ because $P(x)^2=P(Q(x))$=>$2degP=degPdegQ$=>$degQ=2$

Basically same as Iran 3rd round 2017

Yes，in Iran 3rd round,it is even more difficut.