MathDB

1

Part of 2024 ITAMO

Problems(1)

Taking the difference with pi becomes periodic

Source: Italy MO 2024 P1

5/15/2024
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.
algebraalgebra proposedabsolute value