Let $P(x) \in \mathbb{Z}[x]$ be a polynomial. Determine all polynomials $Q(x) \in \mathbb{Z}[x]$, such that for every positive integer $n$, there exists a polynomial $R_n(x) \in \mathbb{Z}[x]$ satisfies $$Q(x)^{2n} - 1 = R_n(x)\left(P(x)^{2n} - 1\right).$$
Problem
Source: Vietnam TST 2024 P6
Tags: algebra, polynomial
math_comb01
27.03.2024 17:10
I assume $P^{2n}$ denotes the power function, not the composition function.
We claim that only $Q = \pm P^k$ works for any positive $k$,
WLOG the initial coefficient of $P,Q$ be positive (else if replace by $-P,-Q$) so for large enough $x$, $P(x)$ and $Q(x)$ both are positive, now fix $x$ and use the following well known result,
Result: If for infinitely many $n$ $a^n-1 \mid b^n-1$ then $b$ is a power of $a$.
Now notice since these are polynomials with bounded degree, there must be some value(as power) that occurs infinitely times, say $k$ then $Q = P^k$
This is a solid overkill, the result used requires very strong theorems to be proven.
Since we have $a^{2n}-1 \mid b^{2n}-1$, we can write this as $(a^2)^n-1 \mid (b^2)^n-1$ now the result is somewhat easy to prove, and is well known,(for eg. see here)
megarnie
27.03.2024 19:34
The official wording had the equation: \[Q(x)^{2n} - 1 = R_n(x) (P(x)^{2n} - 1) \]
kingu
07.07.2024 21:54
I'm curious in what world the gentle finish of the first post is an easy to prove result, lmao.