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Source: San Diego Honors Math Contest 2025 Part II, Problem 2

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somebodyyouusedtoknow

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somebodyyouusedtoknow

#1 h Apr 26, 2025, 11:30 PM • 1 Y

Y by KevinYang2.71

Let $P(x)$ be a polynomial with real coefficients such that $P(x^n) \mid P(x^{n+1})$ for all $n \in \mathbb{N}$ . Prove that $P(x) = cx^k$ for some real constant $c$ and $k \in \mathbb{N}$ .

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#2 h May 8, 2025, 4:12 AM

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Hope it true!
Let $P(x)=cx^k+Q(x)$ with $\deg Q=d <k$
So $cx^{kn}+Q(x^n) \mid cx^{kn+k}+Q(x^{n+1}) \mid x^k \left(cx^{kn}+Q(x^n)\right)+Q(x^{n+1})-x^kQ(x^n)$
This give $cx^{kn}+Q(x^n) \mid Q(x^{n+1})-x^kQ(x^n) \rightarrow dn+d \ge kn, \forall n \ge 0$
$+)$ If $Q \equiv \alpha$
Then, $cx^{kn}+\alpha \mid \alpha -x^k \alpha$ . Thus, $\alpha =0.$ $\blacksquare$
$+)$ If $\deg Q =1$ , then $Q(x)=mx+g$
Then $n+1 \ge kn, \forall n$ so $k=1$ (contradiction). $\blacksquare$
$+)$ If $\deg Q>1$ .
Let $n$ be big enough such that $dn+d>n+k$
Then there exist $R_n(x) \in \mathbb{R}[x]$ with $\deg R=dn+d-kn=l_n$ such that:
$\left(cx^{kn}+Q(x^n) \right)R_n(x)=Q(x^{n+1})-x^kQ(x^n)$ Choose $n$ is even and let $x =-x$ , we have:
$\left(cx^{kn}+Q(x^n) \right)\left(R_n(x)+R_n(-x)\right)=Q(x^{n+1})+Q(-x^{n+1}) (1)$ If $R_n(x)+R_n(-x) \ne 0$ for some $x$
Then $\deg LS \ge kn$ , but $\deg RS \le (d-1)(n+1)$
Choose $n > \frac{d-1}{k+1-d}=n_1$ , we have contradiction.
So that $Q\left(x^{n+1}\right)=-Q\left(-x^{n+1}\right), \forall n >n_1$ is an even number.
Since $x^{2k+1}$ is surjective in $\mathbb{R}$ , we have $Q(x)$ is odd
$\textbf{Claim1:}$ $k$ is even
Let $x=-x$ and $n$ is even in $cx^{kn}+Q(x^n) \mid cx^{kn+k}+Q(x^{n+1})$ , we have: $cx^{kn}+Q(x^n) \mid 2cx^{kn+k}$
Easy to see that $Q(0)=0$ . let $Q(x)=x^sT(x) (T(0) \ne 0)$ . Similary, we have $Q(x) =ax^s$
This give: $cx^{kn}+ax^{sn}\mid ax^{sn+s}-ax^{sn+k}$ or $cx^{kn-sn}+a \mid ax^s-ax^k$ . Choose $n$ is big enough, we have contradiction.
$\textbf{Claim2:}$ $k$ is odd
Then $P(x)$ is odd. So $P(0)=0$ . Let $P(x)=x^uT(x)$ with $T(0) \ne 0$
Easy to see that $T(x^n) \mid T(x^{n+1})$ . Similary, keep it up then $T(0)=0$ (contradiction).
So all solution to this problem are
$\boxed{P(x)=cx^k}$

This post has been edited 5 times. Last edited by luutrongphuc, May 8, 2025, 4:20 AM

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#3 h Thursday at 2:54 PM

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Suppose $r$ is a nonzero root of $P(x)$ . Then any $x_0$ satisfying $x_0^n=r$ is a root of $P(x^n)$ , so it is a root of $P(x^{n+1})$ . Since $x_0^{n+1}=rx_0$ and there are $n$ distinct values of $x_0$ , $x_0^{n+1}$ takes on $n$ distinct values. However, each $x_0^{n+1}$ is a root of $P(x)$ so taking $n$ sufficiently large contradicts the degree of $P$ . Thus $P$ has no nonzero roots so it is in the desired form. $\square$

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