Track / Stage 6 / 222 of 400 #1222 of 1964

Problem 1222

National olympiad, first round

Algebra Difficulty 6.3 Prove it

8-39 Let the non-zero sequence $a_{1}, a_{2}, \cdots$ satisfy: $a_{1}, a_{2}, \frac{a_{1}^{2}+a_{2}^{2}+b}{a_{1} a_{2}}$ are all integers, and
$a_{n+2}=\frac{a_{n+1}^{2}+b}{a_{n}}, n=1,2, \cdots$

where $b$ is a given integer. Prove that every term of the sequence $\left\{a_{n}\right\}$ is an integer.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

[Proof] Let $n \geqslant 2$ , from the recursive formula we get

thus
$b=a_{n+2} a_{n}-a_{n+1}^{2}=a_{n+1} a_{n-1}-a_{n}^{2},$

i.e., $\quad \frac{a_{n+2}+a_{n}}{a_{n+1}}=\frac{a_{n+1}+a_{n-1}}{a_{n}}, n=2,3, \cdots$
Let $c_{n}=\frac{a_{n+2}+a_{n}}{a_{n+1}}$ , then
$c_{n}=c_{1}=\frac{a_{3}+a_{1}}{a_{2}}=\frac{a_{1}^{2}+a_{2}^{2}+b}{a_{1} a_{2}}$

By the assumption, $c_{1}$ is an integer, and
$a_{n+2}=c_{1} a_{n+1}-a_{n}, n=1,2,3, \cdots$

Furthermore, since $a_{1}, a_{2}$ are integers, every term of $\left\{a_{n}\right\}$ is an integer.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.