Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Prove it Saudi Arabia

Consider the sequence $a_1, a_2, a_3, \dots$ defined by $a_1 = 9$ and
$a_{n+1} = \frac{(n+5)a_n + 22}{n+3}$
for $n \ge 1$ . Find all positive integers $n$ for which $a_n$ is a perfect square.

Solution

First of all, we will prove by induction on $n$ that $a_n = n^2 + 7n + 1$ . The base case is trivial. Assume that $a_k = k^2 + 7k + 1$ , so
$a_{k+1} = \frac{(k + 5)a_k + 22}{k + 3} = \frac{(k + 5)(k^2 + 7k + 1) + 22}{k + 3} = (k + 1)^2 + 7(k + 1) + 1$
We distinguish two cases regarding the value of $n$
* if $n \ge 9$ then $(n + 3)^2 < n^2 + 7n + 1 < (n + 4)^2$ , so $a_n$ is not a perfect square.
* if $n \le 8$ then we have

$a_1 = 9$	$a_2 = 31$	$a_5 = 61$	$a_7 = 99$
$a_2 = 19$	$a_4 = 45$	$a_6 = 79$	$a_8 = 121$

Hence,

a_n

is a perfect square if and only if

n = 1, 8

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.