# 1976 USAMO Problems/Problem 5

## Problem

If $P(x)$, $Q(x)$, $R(x)$, and $S(x)$ are all polynomials such that $$P(x^5) + xQ(x^5) + x^2 R(x^5) = (x^4 + x^3 + x^2 + x +1) S(x),$$ prove that $x-1$ is a factor of $P(x)$.

## Solutions

### Solution 1

In general we will show that if $m$ is an integer less than $n$ and $P_0, \dotsc, P_{m-1}$ and $S$ are polynomials satisfying $$P_0(x^n) + x P_1(x^n) + \dotsb + x^{m-1} P_{m-1}(x^n) = \sum_{k=0}^{n-1} x^k S(x),$$ then $P_k(1) = 0$, for all integers $0 \le k \le m-1$. For the problem, we may set $n=5$, $m=3$, and then note that since $P(1)= 0$, $(x-1)$ is a factor of $P(x)$.

Indeed, let $\omega_1, \dotsc, \omega_{n-1}$ be the $n$th roots of unity other than 1. Then for all integers $1 \le i \le n-1$, $$\sum_{k=0}^{m-1} \omega_i^k P_k(\omega_i^n) = \sum_{k=0}^{m-1} \omega_i^k P_k(1) = \sum_{k=0}^{n-1} \omega_i^k S(\omega_i) = \frac{\omega_i^{n-1} - 1}{\omega_i - 1} S(\omega_i) = 0 ,$$ for all integers $1 \le i \le n$. This means that the $(m-1)$th degree polynomial $$\sum_{k=0}^{m-1} x^k P_k(1)$$ has $n-1 > m-1$ distinct roots. Therefore all its coefficients must be zero, so $P_k = 0$ for all integers $0 \le k \le m-1$, as desired. $\blacksquare$

### Solution 2

Let $\zeta, \xi, \rho$ be three distinct primitive fifth roots of unity. Setting $x = \zeta, \xi$, we have \begin{align*} P(1) + \zeta Q(1) + \zeta^2 R(1) &= \frac{\zeta^5-1}{\zeta-1} S(\zeta) = 0, \\ P(1) + \xi Q(1) + \xi^2 R(1) &= \frac{\xi^5-1}{\xi-1} S(\xi) = 0 . \end{align*} These equations imply that $$\zeta Q(1) + \zeta^2 R(1) = \xi Q(1) + \xi^2 R(1),$$ or $$Q(1) = - ( \zeta + \xi)R(1).$$ But by symmetry, $$Q(1) = - (\zeta + \rho)R(1) .$$ Since $\xi \neq \rho$, it follows that $Q(1) = R(1) = 0$. Then, as noted above, $$P(1) = P(1) + \zeta Q(1) + \zeta^2 R(1) = 0,$$ so $(x-1)$ is a factor of $P(x)$, as desired. $\blacksquare$