olymmath

Putnam 2018 B4

  • Session B

Given a real number aa, we define a sequence by x0=1x_0 = 1, x1=x2=ax_1 = x_2 = a, and xn+1=2xnxn−1−xn−2x_{n+1} = 2x_n x_{n-1} - x_{n-2} for n≥2n \geq 2. Prove that if xn=0x_n = 0 for some nn, then the sequence is periodic.

Curated text Kedlaya Putnam archive. Quoted with source.

Solutions

Human solution Solution 1 by Kiran Kedlaya and Lenny Ng

The terms of the source let olymmath link to this solution but not copy it. Read the solution (Kedlaya Putnam archive).

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