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Taking the difference with pi becomes periodic

Source: Italy MO 2024 P1

May 15, 2024
algebraalgebra proposedabsolute value

Problem Statement

Let x0=20242024x_0=2024^{2024} and xn+1=∣xn−π∣x_{n+1}=|x_n-\pi| for n≥0n \ge 0. Show that there exists a value of nn such that xn+2=xnx_{n+2}=x_n.
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