MathDB
Natural Number Sequence

Source: Turkey TST 2014 Day 3 Problem 8

March 12, 2014
number theory proposednumber theory

Problem Statement

a1=5a_1=-5, a2=6a_2=-6 and for all n2n \geq 2 the (an)n=1{(a_n)^\infty}_{n=1} sequence defined as, an+1=an+(a1+1)(2a2+1)(3a3+1)((n1)an1+1)((n2+n)an+2n+1)).a_{n+1}=a_n+(a_1+1)(2a_2+1)(3a_3+1)\cdots((n-1)a_{n-1}+1)((n^2+n)a_n+2n+1)). If a prime pp divides nan+1na_n+1 for a natural number n, prove that there is a integer mm such that m25(modp)m^2\equiv5(modp)