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1976 USAMO Problems/Problem 5

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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 nth 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) = ... 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) ... 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


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1976 USAMO (Problems)
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