We call polynomial $S(x)\in\mathbb{R}[x]$ sadeh whenever it's divisible by $x$ but not divisible by $x^2$. For the polynomial $P(x)\in\mathbb{R}[x]$ we know that there exists a sadeh polynomial $Q(x)$ such that $P(Q(x))-Q(2x)$ is divisible by $x^2$. Prove that there exists sadeh polynomial $R(x)$ such that $P(R(x))-R(2x)$ is divisible by $x^{1401}$.
Problem
Source: Iran MO Third Round A1
Tags: algebra, polynomial
alinazarboland
08.09.2022 19:46
Clearly $P(0)=0$. Now take derivate from that polynomial to get that $P'(0)=2$ since $Q'(0)$ is non-zero. Now we're looking for some $R$ st $R(0)=0$ , $R'(0)$ is non-zero (*), and the polynomial $S(x) = P(R(x))-R(2x)$ satisfies $S^{(i)}(0) = 0 $ for $1 \leq i \leq 1401$ but this is just a system of 1401 linear equation (where each equation has a new variable which is $R^{(i)}(0)$ which will be sth) and we should only fulfill (*) which holds by the fact that $P$ is "sadeh"(which means simple in persian) and we sould've the desired polynomial
topologicalsort
02.01.2025 14:41
Let $Q(x) = x\cdot A(x)$ with $A(0) \neq 0$. Since $P(0) = 0$, let $P(x) = x \cdot T(x)$. We have that $$x \mid A(x)\cdot T(xA(x))-2\cdot A(2x)$$so $T(0) =2$ and let $T(x) = 2+x\cdot B(x)$. Now the expression becomes $$2\cdot A(x)+x\cdot A(x)^2\cdot B(xA(x))-2\cdot A(2x)$$We will inductively constuct $A_n(x)$ such that $\operatorname{deg}A_n(x) < n$ and $$x^n \mid 2\cdot A_n(x)+x\cdot A_n(x)^2\cdot B(xA_n(x))-2\cdot A_n(2x)$$for all $n$. We can set $A_1(x) = c$ for some non-zero $c$ and assume we have constructed $A_n(x)$. Let $A_{n+1}(x) = a_nx^n+A_n(x)$. The coefficient of $x^n$ in the expression is $$2a_n+C-2 \cdot 2^na_n$$where $C$ is the coefficient of $x^n$ in $x\cdot A_{n+1}(x)^2\cdot B(xA_{n+1}(x))$ which does not depend on $a_n$. Now $a_n = \frac{C}{2(2^n-1)}$ is enough for $A_{n+1}$ to satisfy the condition. Finally $R(x) = x\cdot A_{1500}(x)$ finishes the problem.