HMMT-Feb 2022 alg, Problem 7

Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right),\left(x_{4}, y_{4}\right)$, and $\left(x_{5}, y_{5}\right)$ be the vertices of a regular pentagon centered at $(0,0)$. Compute the product of all positive integers $k$ such that the equality \[ x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k}+x_{5}^{k}=y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}+y_{5}^{k} \] must hold for all possible choices of the pentagon.

Solution: Solution: Without loss of generality let the vertices of the pentagon lie on the unit circle. Then, if $f(\theta)=\cos (\theta)^{k}$ and $g(\theta)=\sum_{j=0}^{4} f(\theta+2 j \pi / 5)$, the condition becomes $g(\theta)=g(\pi / 2-\theta)$, or $g(\theta)=$ $g(\theta+\pi / 2)$, since $g$ is an odd function. Write $f \asymp g$ if $f=c g$ for some nonzero constant $c$ that we don't care about. Since $\cos \theta \asymp e^{i \theta}+e^{-i \theta}$, we find that \[ f(\theta) \asymp \sum_{\ell \in \mathbb{Z}}\left(\begin{array}{c} k \\ \frac{k+\ell}{2} \end{array}\right) e^{i \ell \theta} \] where $\left(\begin{array}{l}a \\ b\end{array}\right)$ is defined to be zero if $b$ is not an integer in the interval $[0, a]$. It is also true that \[ \sum_{j=0}^{4} e^{i \ell(\theta+2 j \pi / 5)}= \begin{cases}5 e^{i \theta} & 5 \mid \ell \\ 0 & \text { else }\end{cases} \] so \[ g(\theta) \asymp \sum_{\ell \in 5 \mathbb{Z}}\left(\begin{array}{c} k \\ \frac{k+\ell}{2} \end{array}\right) e^{i \ell \theta} \] This is periodic with period $\pi / 2$ if and only if all terms with $\ell$ not a multiple of 4 are equal to 0 . However, we know that the nonzero terms are exactly the $\ell$ that (1) are multiples of $5,(2)$ are of the same parity as $k$, and (3) satisfy $|\ell| \leq k$. Hence, if $k$ is even, the condition is satisfied if and only if $k<10$ (else the $\ell=10$ term is nonzero), and if $k$ is odd, the condition is satisfied if and only if $k<5$ (else the $\ell=5$ term is nonzero). Our final answer is $1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 8=1152$. $\fbox{1152}$.

Answer: 1152