Is this From the Romanian Math Masters ?

@above. See here for the information.

Proposed by Navid Safaie (I.R.Iran)

To any linear map $f$ given by $t\to at+b$ assign the the linear polynomial $f(x,y)=x-ay-b.$ Claim 1. If $f(x,y)$ and $g(x,y)$ are linear factors of $Q$, then so is $f\circ g(x,y).$ Proof: Just use the fact that $Q(f\circ g(x), g(x))+Q(g(x),x)=0.$ $\blacksquare$ Observation 1. If $f\circ g$ is a constant transformation, then $f$ is constant, or $g$ is constant. Observation 2. If $f(x,y)$ is a linear factor of $Q$, then $f$ is not constant. Let $G$ be the set of all linear factors $f(x,y)$ of $Q.$ Observations 1,2 along with Claim 1 and the finiteness of $G$ are easily seen to imply that $(G,\circ)$ is a group with identity element $\text{id}(x,y)=x-y.$ Since $|G|=k$, Lagrange's theorem implies that $f^k=\text{id}$ for any $f(x,y)\in G.$ Let $f_i(x)=x-a_iy-b_i,a_i\neq 0$ for $1\le i\le k$ be the linear factors of $Q$. The above tells us that $a_i$ are $k$-th roots of unity. Claim 2. If $i\neq j$, then $a_i\neq a_j$. Proof: If $a_i=a_j$, then $P(x)=P(f_i^{-1}(x))=P(f_j(f_i^{-1}(x)))=P(x+b_j-b_i)$, and since $P$ is not constant, $b_i=b_j$, i.e. $f_i=f_j$, contradicition! Claim 2 gives us that the $a_i$'s are all $k$-th roots of unity, implying that $G$ is cyclic and WLOG $f_i(x,y)=x-a_1^{i}y-\frac{a_1^{i}-1}{a_1-1}b_1$. Overall, $$\prod_{i=1}^{k}f_i(x,y)=\left(x+\frac{b_1}{a_1-1}\right)^k-\left(y+\frac{b_1}{a_1-1}\right)^k|Q(x,y).$$By shifting $P$, WLOG $x^k-y^k|P(x)-P(y).$ Now, there is a polynomial $S$ such that $P(x)=S(x^k)$, and, assuming that $R$ is not a product of linear factors, WLOG $R|T(x,y)=\frac{S(x^k)-S(y^k)}{x^k-y^k}.$ Let $m$ be the degree of $S$ and write $T=RU$, $R$ of degree $a$, $U$ of degree $b$. Let $R_d,U_d$ be the sum of terms of degree $d$ in $R$ and $U$, respectively. Then $R_a(x,y)\cdot U_b(x,y)=c\cdot\cdot\frac{x^{mk}-y^{mk}}{x^k-y^k}=\text{product~of~distinct~linear~factors}$, meaning that $(R_a,U_b)=1.$ Claim 3. $R_d=U_e=0$ for any $d<a$ and $b>e\ge b-a.$ Proof: Suppose that we have proven that $R_{a-1}=U_{b-1}=R_{a-2}=U_{b-2}=...=R_{a-i}=U_{b-i}=0.$ Then $$R_aU_{b-i-1}+U_{b}R_{a-i-1}=\text{sum~of~terms~of~degree}~a+b-i-1.$$However, $a+b-(i-1)=mk-(i-1)\ge mk-a>(m-1)k$, i.e. $R_aU_{b-i-1}+U_{b}R_{a-i-1}=0$. $(R_a,U_b)=1$ yields $R_a|R_{a-i-1}$ and degree bounding gives $R_{a-i-1}=0$ and then $U_{b-i-1}=0$ follows as well. It remains to prove that $R_{a-1}=U_{b-1}=0$, but this can be done in the same fashion.$\blacksquare$ We have $R\cdot U_b=R_a\cdot U_b=\text{product~of~linear~factors}$, implying that $R$ is a product of linear factors.

Interesting to see two RMM problems involving chains and polynomials to be placed as P6. First things first, if $x-ay-b\mid P(x)-P(y)$ then $P(y)=P(ay+b) \forall y\in\mathbb{C}$ Let $f(x)=ax+b$. Then, for some $t$, $f^t(x)=x$, as $f$ is bijective and there doesn’t exist a chain of path of length $>\deg P$. Claim: there exist $a_0, b_0$ such that if $P(x)\equiv P(g(x))$ for some linear function $g$ then $g=f^k $ for some $k\in\mathbb{Z}$ Proof: Note if $f(x)=ax+b$ has $f^t(x)=x$ $a$ is a root of unity. Use Bezout to win. Let $T(x)=P(x-\frac{b}{a-1})$ then observe $T(x)=T(ax)$. Let $n$ be the smallest number such that $a^n=1$. Algebraic manipulations yield $T(x)=S(x^n)$ for some $S(x)\in \mathbb{C}[x]$ It remains to show if $R(x,y)\mid T(x)-T(y)$ and $\deg R\le t$ then $R$ can be factored into linear factors, as a transformation would bring it back to $P$. If $M(x,y)=S(x)-S(y)$ has a linear factor other than $x-y$, then $T$ has linear factors not dividing $x^n-y^n $, absurd. Therefore $R(x,y)\mid \frac{S(x^n)-S(y^n)}{x^n-y^n} \prod\limits_{e=0}^{n-1} (x-a^ey)$ Removing linear factors, we can see that $R'(x,y)\mid \frac{S(x^n)-S(y^n)}{x^n-y^n}$ If $n=1$ we are trivially done, so we assume $n>1$. Let $R'(x,y)A(x,y)=\frac{S(x^n)-S(y^n)}{x^n-y^n}$. Let $r,a$ denote the degrees of $R',A$ and let $R_m$ be the sum of degree $m$ terms of $R'(x,y)$ and define $A_m$ similarly. Note $n\mid r+a$, so $R_rA_a=c\frac{x^{r+a}-y^{r+a}}{x^n-y^n}$. Since $R_r\ne 0$, $R_r$ is the product of $r$ $(x-\omega y)$ where $\omega$ is an $(r+a)$th root of unity but not an $\frac{r+a}{n}$th RoU, and $A_m$ contains the other factors. Since $R_rA_{a-1} = -R_{r-1}A_a$ as $n\nmid r+a-1$, we can see that $R_{r-1}$ must be divisible by $R_r$ as $\gcd(A_a,R_r)$ is a constant. This implies $R_{r-1}=A_{a-1}=0$. We can repeat the same argument to show $R_{r-2}=\cdots=R_1=R_0=0$, so $R(x,y)=R_r$ is a product of linear factors.

Is this really from RMM?

huh what do you mean @above

To clarify, this is the original RMM #6. The original day 2 paper was leaked to Bulgaria in day 1 by mistake, so those problems were not in the exam and are in the shortlist instead.

Like the solutions above,we prove that $P(x)=Q(x^k)$.Now suppose that there is an irriductable polyominial $R(x,y)$ of degree $t<k$ dividing $Q(x,y)$.Clearly $R$ is not homogenous(otherwise it must be linear).Denote by $w$ the $k-th$ root of unity.We have $R(w^ax,w^by)|Q(w^ax,w^by)=Q(x,y)$.By comparing the lowest and highest degree term of $R(w^ax,w^by)$ one see that these $k^2$ polyominials are distinct,therefore their product $M(x,y)$ divides $Q(x,y)$.Now,comparing the highest degree terms.It is easy to see that $M(x,y)$ is a $t-th$ power of another polyominial,while the highest term of $Q(x,y)$ is like $x^n-y^n$,having no double roots.This immediately yields that $t=1$.

, can someone please tell why $M(x,y)$ is a $t-th$ power of another polynomial ?