MathDB
2021 Algebra/NT #8: LCM's and GCD's

Source:

May 30, 2021
number theory

Problem Statement

For positive integers aa and bb, let M(a,b)=lcm(a,b)gcd(a,b),M(a,b) = \tfrac{\text{lcm}(a,b)}{\gcd(a,b)}, and for each positive integer n2,n \ge 2, define xn=M(1,M(2,M(3,,M(n2,M(n1,n))))).x_n = M(1, M(2, M(3, \dots , M(n - 2, M(n - 1, n))\cdots))). Compute the number of positive integers nn such that 2n20212 \le n \le 2021 and 5xn2+5xn+12=26xnxn+1.5x_n^2 + 5x_{n+1}^2 = 26x_nx_{n+1}.