HMMT February

Answer: $\quad P(x)=\frac{x\left(x^{4 k+2}+1\right)}{x^{2}+1}$ or $P(x)=\frac{x\left(1-x^{4 k}\right)}{x^{2}+1}$ Solution: It is straightforward to plug in and verify the above answers. Hence, we focus on showing that these are all possible solutions. The key claim is the following. Claim: If $r \neq 0$ is a root of $P(z)$ with multiplicity $n$, then $1 / r$ is also a root of $P(z)$ with multiplicity $n$. Proof 1 (Elementary). Let $n^{\prime}$ be the multiplicity of $1 / r$. It suffices to show that $n \leq n^{\prime}$ because we can apply the same assertion on $1 / r$ to obtain that $n^{\prime} \leq n$. To that end, suppose that $(z-r)^{n}$ divides $P(z)$. From the equation, we have $$z^{N}\left[P\left(\frac{1}{z}\right)+P(z)\right]=z^{N}\left[\left(z+\frac{1}{z}\right) P(z) P\left(\frac{1}{z}\right)\right]$$ where $N \gg \operatorname{deg} P+1$ to guarantee that both sides are polynomial. Notice that the factor $z^{N} P(z)$ and the right-hand side is divisible by $(z-r)^{n}$, so $(z-r)^{n}$ must also divide $z^{N} P\left(\frac{1}{z}\right)$. This means that there exists a polynomial $Q(z)$ such that $z^{N} P\left(\frac{1}{z}\right)=(z-r)^{n} Q(z)$. Replacing $z$ with $\frac{1}{z}$, we get $$\frac{P(z)}{z^{N}}=\left(\frac{1}{z}-r\right)^{n} Q\left(\frac{1}{z}\right) \Longrightarrow P(z)=z^{N-n}(1-r z)^{n} Q\left(\frac{1}{z}\right)$$ implying that $P(z)$ is divisible by $(z-1 / r)^{n}$. Proof 2 (Complex Analysis). Here is more advanced proof of the main claim. View both sides of the equations as meromorphic functions in the complex plane. Then, a root $r$ with multiplicity $n$ of $P(z)$ is a pole of $\frac{1}{P(z)}$ of order $n$. Since the right-hand side is analytic around $r$, it follows that the other term $\frac{1}{P(1 / z)}$ has a pole at $r$ with order $n$ as well. By replacing $z$ with $1 / z$, we find that $\frac{1}{P(z)}$ has a pole at $1 / r$ of order $n$. This finishes the claim. The claim implies that there exists an integer $k$ and a constant $\epsilon$ such that $$P(z)=\epsilon z^{k} P\left(\frac{1}{z}\right)$$ By replacing $z$ with $1 / z$, we get that $$z^{k} P\left(\frac{1}{z}\right)=\epsilon P(z)$$ Therefore, $\epsilon= \pm 1$. Moreover, using the main equation, we get that $$\frac{1}{P(z)}+\frac{\epsilon z^{k}}{P(z)}=z+\frac{1}{z} \Longrightarrow P(z)=\frac{z\left(1+\epsilon z^{k}\right)}{1+z^{2}}$$ This is a polynomial if and only if $(\epsilon=1$ and $k \equiv 2(\bmod 4))$ or $(\epsilon=-1$ and $k \equiv 0(\bmod 4))$, so we are done.