AoPS - Problem

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Problem

Source: Vietnam TST 2023 P4

Tags: number theory

a_507_bc

14.04.2023 10:06

Given are two coprime positive integers $a, b$ with $b$ odd and $a>2$ . The sequence $(x_n)$ is defined by $x_0=2, x_1=a$ and $x_{n+2}=ax_{n+1}+bx_n$ for $n \geq 1$ . Prove that: $a)$ If $a$ is even then there do not exist positive integers $m, n, p$ such that $\frac{x_m} {x_nx_p}$ 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 $\frac{x_m} {x_nx_p}$ is a perfect square.

Plimpton322

14.04.2023 16:28

Use $p-$ adic valuation.

Plimpton322

14.04.2023 20:01

$v_2(x_n)=\begin{cases}1,\quad 2\mid n\\ v_2(a),\quad 2\nmid n.\end{cases}$

$v_2(x_n)=\begin{cases}1,\quad n\equiv 0\pmod{6}\\ v_2(a^2+3b),\quad n\equiv 3\pmod{6}\\ 0,\quad n\not\equiv 0\pmod{3}. \end{cases}$

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?