MathDB
b_1 = a_0, b_{n+1} = P (b_n) , P (x) = a_dx^d + ... + a_2x^2 + a_0

Source: 2021 Saudi Arabia Training Lists p34 https://artofproblemsolving.com/community/c2758131_2021_saudi_arabia_training_tests

January 5, 2022
polynomialnumber theorydivides

Problem Statement

Let coefficients of the polynomialP(x)=adxd+...+a2x2+a0 P (x) = a_dx^d + ... + a_2x^2 + a_0 where d2d \ge 2, are positive integers. The sequences (bn)(b_n) is defined by b1=a0b_1 = a_0 and bn+1=P(bn)b_{n+1} = P (b_n) for n1n \ge 1. Prove that for any n2n \ge 2, there exists a prime number pp such that pbnp|b_n but it does not divide b1,b2,...,bn1b_1, b_2, ..., b_{n-1}.