lemma1:$\forall x\in \mathbb{\mathbb{C}}\rightarrow q(x)\in \mathbb{R}$ then $q(x)=a\in \mathbb{R}$ . lemma2:$\forall x\in \mathbb{\mathbb{C}}$ if $p(x)\in \mathbb{R}\rightarrow q(x)\in \mathbb{R}$ then there exist $s(x)\in \mathbb{R}[x])$ such that q(x)=s(p(x)). you can see in initial problem $p(x)\in \mathbb{R}\leftrightarrow q(x)\in \mathbb{R}$ .the rest is easy.

Consider $p(z) = (1 + i)z$ and $q(z) = (1 - i)z$. Then clearly $p(z)q(\overline{z}) = (1 + i)(1 - i)z\overline{z} \in\mathbb{R}$ for every $z$. However, $q(z) \ne 0$ and $p(z) \ne q(z) k$ for any real number $k$. ($p(z) = iq(z)$, but $i$ is certainly not a real number). However, we do notice that $p(z) = k\overline{q}(z)$ where $k = 1$ here. And in fact, this can be proven. Write \[ p(z)q(\overline{z}) = \frac{p(z) q(\overline{z})\overline{q}(z)}{\overline{q}(z)} \] since we know this whole quantity is real and $q(\overline{z})\overline{q}(z)$ is real, we know that $f(z) = \frac{p(z)}{\overline{q}(z)}$ is real for every $z$. But since $p$ and $\overline{q}$ are polynomials we can find some ball where $f(z)$ is holomorphic, and so by the open mapping theorem we conclude that $f(z)$ is a constant as desired.

devenware wrote: Consider $p(z) = (1 + i)z$ and $q(z) = (1 - i)z$. Then clearly $p(z)q(\overline{z}) = (1 + i)(1 - i)z\overline{z} \in\mathbb{R}$ for every $z$. However, $q(z) \ne 0$ and $p(z) \ne q(z) k$ for any real number $k$. ($p(z) = iq(z)$, but $i$ is certainly not a real number). However, we do notice that $p(z) = k\overline{q}(z)$ where $k = 1$ here. And in fact, this can be proven. Write \[ p(z)q(\overline{z}) = \frac{p(z) q(\overline{z})\overline{q}(z)}{\overline{q}(z)} \] since we know this whole quantity is real and $q(\overline{z})\overline{q}(z)$ is real, we know that $f(z) = \frac{p(z)}{\overline{q}(z)}$ is real for every $z$. But since $p$ and $\overline{q}$ are polynomials we can find some ball where $f(z)$ is holomorphic, and so by the open mapping theorem we conclude that $f(z)$ is a constant as desired. sorry in original problem $p,q\in \mathbb{R}[x]$ i forgot to write that.i fixed it

@mmaht, ah, I'm glad there wasn't an error on the Iranian Olympiad! Also, to anyone else reading, note that my solution implies the result in the current version of the problem, since $\overline{q}(z) = q(z)$!

mmaht wrote: lemma1:$\forall x\in \mathbb{\mathbb{C}}\rightarrow q(x)\in \mathbb{R}$ then $q(x)=a\in \mathbb{R}$ . Sorry, I don't understand this lemma. For example if $q(x)=x^2+1$ then both $q(i)$ and $q(2i)$ are reals but clearly they aren't the same real. @EDIT: sorry, what a silly mistake

But $q(1+i)\not\in \mathbb{R}$

First prove that degP=degQ .( Use $p(z)q(\overline{z})$=$q(z)p(\overline{z})$ . then use induction on (degP+degQ).

with the conditions which have been mentioned for the problem we have that: lemma: if $p(z)q(\overline{z}) \in \mathbb{R}$ and $q(x)$ isn't zero the $deg P=deg Q$ in which can be proved using induction as we have $p(z)q(\overline{z})=q(z)p(\overline{z})$ ! now if $p(x)=a_nx^n+...+a_1x+a_0$ and $q(x)=b_nx^n+...+b_1x+b_0$ then because $a_0, b_0$ cannot be zero then we shall put both of them equal to one! and now note $p(x)-q(x)$ and $p(x)$ the condition of problem will be true so according to the lemma because $deg p(x) \ge deg(p(x)-q(x))$ then $p(x)-q(x)=0 $ Q.E.D

$p(z)q(\overline{z}) \in \mathbb{R}$ so $\overline{p(z)q(\overline{z})} \in \mathbb{R} \implies q(z)p(\overline{z}) \in \mathbb{R}$ which implies $\deg p = \deg q$. Assume that $p(x) = a_nx^n + ... + a_0$ and $q(x) = b_nx^n + ... + b_0$. By Induction assume that $p(x) = kq(x)$ for $\deg p = \deg q = n$. we will prove that for $p(x) = sx^{n+1} + a_nx^n + ... + a_0$ and $q(x) = tx^{n+1} + b_nx^n + ... + b_0$ such that suit the problem condition then we have $\frac{s}{t} = k$. $p(x)q(\overline{x}) = stx^{n+1}\overline{x^{n+1}} + sx^{n+1}\overline{(b_nx^n + ... + b_0)} + t\overline{x^{n+1}}(a_nx^n + ... + a_0) + (a_nx^n + ... + a_0)(\overline{b_nx^n + ... + b_0})$ Note that $stx^{n+1}\overline{x^{n+1}} , (a_nx^n + ... + a_0)(\overline{b_nx^n + ... + b_0})\in \mathbb{R}$ so for $z = c + di$ we need to have $sx^{n+1}\overline{(b_nx^n + ... + b_0)} + t\overline{x^{n+1}}(a_nx^n + ... + a_0) \in \mathbb{R}$ since $\overline{z} = c - di$ so after expanding $sx^{n+1}\overline{(b_nx^n + ... + b_0)} + t\overline{x^{n+1}}(a_nx^n + ... + a_0)$ it's in form of $e + fi$ where $f$ is $0$ cause it's real so after checking Coefficients of $i$ (it's large so you can check it for smaller cases like $n = 0,1,2$ to understand) we will conclude that $\frac{s}{t} = \frac{a_i}{b_i}$ for $1 \le i \le n$. case $n = 0$: $p(x) = a_1x + a_0$ and $q(x) = b_1x + b_0$ we need to have $a_1b_0x + a_0b_1\overline{x}\in \mathbb{R}$ or $(a_1b_0c + a_0b_1c) + (a_1b_0d - a_0b_1d)i \in \mathbb{R}$ or $(a_1b_0d - a_0b_1d) = 0$ which implies $a_1b_0 = a_0b_1$ or $\frac{a_1}{b_1} = \frac{a_0}{b_0}$.

Mahdi_Mashayekhi wrote: $p(z)q(\overline{z}) \in \mathbb{R}$ so $\overline{p(z)q(\overline{z})} \in \mathbb{R} \implies q(z)p(\overline{z}) \in \mathbb{R}$ which implies $\deg p = \deg q$. . Sorry, i dont understand, can you explain this ? Thanks.