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Question : Supposing that a sequence $\{a_n\}$ is defined as $${a_{n+3}}^2=-{a_{n+2}}^2+2{a_{n+1}}^2+48a_{n+1}a_{n}+32{a_n}^2\ (n\ge 1)$$ $$a_1=a_2=a_3=1$$ then, is $a_n$ a square number for any $n$?

For example, we can see $$\sqrt{a_n} : 1,1,1,3,1,5,7,3,17,11,23,45,1,91,89,93,271,85,457,627,287,1541,967,2115,\cdots$$

Motivation : I found the following question in a book without any proof.

Supposing that a sequence $\{b_n\}$ is defined as $$b_{n+3}=-b_{n+2}+2b_{n+1}+8b_n\ (n\ge 1)$$ $$b_1=b_2=b_3=1,$$ then, prove that $b_n$ is a square number for any $n$.

This is obvious by the following relational expression : $$(b_{n+3}-b_{n+2})^2=64b_{n+1}b_n,$$ which can be shown by induction on $n$.

After solving this question, I've tried to find a similar sequence by using computer. Then, I reached the above expectation. The expectation seems true, but I can neither find any counterexample nor prove that the sequence always gives a square number. Can anyone help?

mathlove
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  • which book? Thank you,and your all problem from which book?Thank you –  Oct 03 '13 at 09:32
  • @nanchangjian: Sadly I'm not sure its title. This is because I like borrowing many books from a library and noting what interests me without noting their titles. Anyway, in future I'm going to write books' titles. Thank you for pointing that out. By the way, note that I don't write "found in a book" in 'all' of my questions. Be careful about the words you use. – mathlove Oct 03 '13 at 12:32
  • I would be pretty amazed that you found this question by using a computer search. Motivation for it stems partly from these sequences in OEIS. – Calvin Lin Oct 03 '13 at 14:01
  • @CalvinLin: well, to be honest, I've spent a lot of time and I was very surprised to get this one! – mathlove Oct 03 '13 at 14:30
  • @mathlove Cool! That's pretty exciting! There are several math theorems that arise that way too, though you should look into why these things happen. IE most of the time, it's not by chance, but there is a deeper meaning to it. – Calvin Lin Oct 03 '13 at 22:58

1 Answers1

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Let $b_n = -b_{n-1} - 2b_{n-2}$, $b_1=1, b_2=-1$.

I claim that $a_n = b_n^2$, which is obviously a perfect square, for all $n \in \Bbb{Z}^+$. (See footnote below)

Proof by induction:

Base case: $a_1 = a_2 = a_3 = b_1^2 = b_2^2 = b_3^2 = 1$

Hypothesis: $a_n = b_n^2$ for $n \le k$

Induction step: Start by using the induction hypothesis on the expression for $a_{k+1}$:

$$\begin{align} &a_{k+1}^2 = -a_k^2+2a_{k-1}^2+48a_{k-1}a_{k-2}+32a_{k-2}^2 = -b_k^4 + 2b_{k-1}^4+48b_{k-1}^2b_{k-2}^2+32b_{k-2}^4 \end{align}$$

We want to show that $a_{k+1}^2 = b_{k+1}^4$. Expanding $b_{k+1}^4$ gives:

$$\begin{align} &b_{k+1}^4 = (-b_k - 2b_{k-1})^4 = b_k^4 + 8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+16b_{k-1}^4 = \\\\ &-b_k^4+2b_{k-1}^4+48b_{k-1}^2b_{k-2}^2+32b_{k-2}^4 + \\ &{\color{red}{2b_k^4+8b_k^3b_{k-1}+24b_k^2b_{k-1}^2+32b_kb_{k-1}^3+14b_{k-1}^4-48b_{k-1}^2b_{k-2}^2-32b_{k-2}^4}} \end{align}$$

If we now can show that the red part equals zero, we're done. This can be done by replacing $b_k$ by $(-b_{k-1}-2b_{k-2})$ and expanding.

Motivation for my claim: A search on OEIS.

EDIT: I just realized that proof by induction is completely unnecessary. Showing that the expansion of $b_{n}^4$ is equal to the recursion formula for $a_n^2$ is enough, i.e. the details in the induction step above. I guess I was too involved in the algebraic manipulations to see the big picture at the time.

Daniel R
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