Given a positive integer $k$, find all polynomials $P$ of degree $k$ with integer coefficients such that for all positive integers $n$ where all of $P(n)$, $P(2024n)$, $P(2024^2n)$ are nonzero, we have $$\frac{\gcd(P(2024n), P(2024^2n))}{\gcd(P(n), P(2024n))}=2024^k.$$Allen Wang
Problem
Source: ELMO Shortlist 2024/N3
Tags: number theory, Elmo, ratio
MarkBcc168
22.06.2024 18:50
The answer is $P(x) =ax^k$ for any $a\in\mathbb Z$, which obviously works.
Pick a positive integer $n$ greater than all roots of $P$. Then, note that for all positive integers $m$, we have
\begin{align*}
&\frac{\gcd(P(2024^{m}n), P(2024^{m+1}n))}
{\gcd(P(2024^{m-1}n), P(2024^{m}n))} = 2024^k \\
\stackrel{\text{induction}}{\implies}\quad
&\frac{\gcd(P(2024^{m}n), P(2024^{m+1}n))}
{\gcd(P(n), P(2024n))} = 2024^{km},
\end{align*}so $2024^{km} \mid P(2024^m n)$. However, since $\deg P=k$, the sequence $x_m = \tfrac{P(2024^m n)}{2024^{km}}$ is bounded, so by pigeonhole,
there is a constant $c$ and infinitely many $m$ such that $P(2024^m n) = c\cdot 2024^{km}$. This is enough to imply that $P(x) = ax^k$, where $a = c/n^k$.
P2nisic
23.06.2024 16:18
IAmTheHazard wrote: Given a positive integer $k$, find all polynomials $P$ of degree $k$ with integer coefficients such that for all positive integers $n$ where all of $P(n)$, $P(2024n)$, $P(2024^2n)$ are nonzero, we have $$\frac{\gcd(P(2024n), P(2024^2n))}{\gcd(P(n), P(2024n))}=2024^k.$$Allen Wang Let $P(x)=x^aQ(x)=x^a[b_{k-a}x^{k-a}+...+b_1x+b_0$ then we have: $2024^k=\frac{\gcd(P(2024n), P(2024^2n))}{\gcd(P(n), P(2024n))}=\frac{2024^an^a(Q(2024n),2024^aQ(2024^2n))}{n^a(Q(n),2024^aQ(2024n))}=2024^a\frac{(Q(2024n),2024^aQ(2024^2n))}{(Q(n),2024^aQ(2024n))}$ Let $U_{2024}(b_0)=c$ if we take $n=2024^{c+1}$ we have that: $U_{2024}(Q(n))=U_{2024}(Q(2024n))=U_{2024}(Q(2024^2n))=c$ and so we have that: $U_{2024}(2024^k)=U_{2024}(2024^a\frac{(Q(2024n),2024^aQ(2024^2n))}{(Q(n),2024^aQ(2024n))})=a+c-c=a$ So we get that $k=a$ and that $P(x)=ax^k$ wgich works