Solutions to the 78th William Lowell Putnam Mathematical Competition Saturday, December 2, 2017
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Remark: A similar argument shows that any secondorder linear recurrent sequence also satisfies a quadratic second-order recurrence relation. A familiar example is the identity Fn−1Fn+1 − F2 n = (−1)n for Fn the nth Fibonacci number. More examples come from various classes of orthogonal polynomials, including the Chebyshev polynomials mentioned below. Second solution. We establish directly that Qn(x) = xQn−1(x)−Qn−2(x), which again suffices. From the equation
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