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Let $(a_n)$ be the sequence of integers defined recursively by $a_1 = a_2 = 1$ and $a_{n+2}=7a_{n+1}-a_{n}-2$ for $n \ge 1$. Prove that $a_n$ is a perfect square for every n.

I found this question in The IMO Compendium (longlisted problem 64 of IMO 1986)

I began by listing the terms as follows

$a_3 = 5^2$

$a_4 = 13^2$

$a_5 = 34^2$

$a_6 = 89^2$

$a_7 = 233^2$

Taking difference of the terms which are squared gives another sequence $(8, 21, 55, 144....)$

However I don't see the pattern in either of these. It would be appreciated if someone could help me in finding general term of these sequences, from where I believe it can be proved fairly easily.

If there is an alternate approach to this prove, even that would help.

Centelle
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1 Answers1

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Let us define the sequence $b_n$as:

$b_1=1,b_2=1,b_{n+2}=3b_{n+1}-b_{n}$

Claim) $b_n^2 + b_{n+1}^2 = 3b_{n+1}b_n - 1$ for integer $n$.

Can be easily proved by induction on $n$.

Claim) $a_n=b_n^2$ for all positive integer $n$.

$7b_{n+1}^2-b_n^2-2=9b_{n+1}^2+b_n^2-2(b_{n+1}^2+b_n^2+1)$

$=9b_{n+1}^2-6b_nb_{n+1}+b_n^2=(3b_{n+1}-b_n)^2=b_{n+2}^2$.

Also note that $b_1^2=a_1=1$ and $b_2^2=a_2=1$.

So, $b_n^2$ and $a_n$ are same sequences.

Therefore, $a_n=b_n^2$, so $a_n$is a perfect square for all n.

HJS
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  • different emphasis, when $x_{n+2} = C x_{n+1} - x_n,$ then $x_{n+2}^2 - (C^2-2)x_{n+1}^2 + x_n^2$ is constant as $n$ changes. – Will Jagy Oct 02 '21 at 17:59