Given are two coprime positive integers a,b with b odd and a>2. The sequence (xn) is defined by x0=2,x1=a and xn+2=axn+1+bxn for n≥1. Prove that: a) If a is even then there do not exist positive integers m,n,p such that xmxnxp is a positive integer. b) If a is odd then there do not exist positive integers m,n,p such that mnp is even and xmxnxp is a perfect square.
Problem
Source: Vietnam TST 2023 P4
Tags: number theory
Plimpton322
14.04.2023 16:28
Use p−adic valuation.
Plimpton322
14.04.2023 20:01
v2(xn)={1,2∣nv2(a),2∤
v_2(x_n)=\begin{cases}1,\quad n\equiv 0\pmod{6}\\ v_2(a^2+3b),\quad n\equiv 3\pmod{6}\\ 0,\quad n\not\equiv 0\pmod{3}. \end{cases}
HaO-R-Zhe
14.04.2023 20:30
Plimpton322 wrote:
v_2(x_n)=\begin{cases}1,\quad 2\mid n\\ v_2(a),\quad 2\nmid n.\end{cases}
I don't see how to finish from here? We may allow a=4 and clearly, \nu_2(x_n) = 2 for n odd. One may choose say a odd n and \frac{x_n}{x_{n-1}x_2} and \nu_2 works out.
Tuan.NT
18.04.2023 03:54
b) Assume that m,n and p are positive integer numbers such that mnp is even and x_m/x_nx_p is a square. Then x_n|x_m and x_p|x_m, therefore by a well-known result, m/n and m/p are odd integers, hence m,n and p are even integers. Now we can have the Jacobi symbol (2|D)=-1, a contradiction. Remark: See in Colloquium Mathematicum, Vol. 130, No. 1, 2013.
Jjesus
07.09.2023 22:54
Any solution?