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0113 number theory 1st edition Round 1 p3

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May 9, 2021
number theory1st edition

Problem Statement

Let x0=1x_0 = 1 and x1=2003x_1 = 2003 and define the sequence (xn)n≥0(x_n)_{n \ge 0} by: xn+1=xn2+1xn−1x_{n+1} =\frac{x^2_n + 1}{x_{n-1}} , ∀n≥1\forall n \ge 1 Prove that for every n≥2n \ge 2 the denominator of the fraction xnx_n, when xnx_n is expressed in lowest terms is a power of 20032003.
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