MathDB
Integer sequence

Source: VMO 2019

April 15, 2019
Integer sequenceperfect numbersum of divisors

Problem Statement

Let (xn)({{x}_{n}}) be an integer sequence such that 0x0<x11000\le {{x}_{0}}<{{x}_{1}}\le 100 and xn+2=7xn+1xn+280, n0.{{x}_{n+2}}=7{{x}_{n+1}}-{{x}_{n}}+280,\text{ }\forall n\ge 0. a) Prove that if x0=2,x1=3{{x}_{0}}=2,{{x}_{1}}=3 then for each positive integer n,n, the sum of divisors of the following number is divisible by 2424 xnxn+1+xn+1xn+2+xn+2xn+3+2018.{{x}_{n}}{{x}_{n+1}}+{{x}_{n+1}}{{x}_{n+2}}+{{x}_{n+2}}{{x}_{n+3}}+2018. b) Find all pairs of numbers (x0,x1)({{x}_{0}},{{x}_{1}}) such that xnxn+1+2019{{x}_{n}}{{x}_{n+1}}+2019 is a perfect square for infinitely many nonnegative integer numbers n.n.