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Can anyone give me a hint on this?

Is there a sequence of positive integers such that $(a_{n+3}-a_{n+2})^2=a_{n+1}+a_n$ for all $n$? Or strongly, $a_{n+3}-a_{n+2}=\sqrt{a_{n+1}+a_n}$.

If there is, how can I find it?

If there is not, how can I prove it?

About the erased comment and to prevent misunderstandings: The sequence $a_n=0$ for all $n$ does not work since $0$ is not positive.

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  • $\begingroup$ $0$ is not positive. $\endgroup$ – Ury May 22 '16 at 6:31
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    $\begingroup$ $1,3,1,3,1,3,1,3\dots$ $\endgroup$ – almagest May 22 '16 at 6:34
  • $\begingroup$ Oh, thanks. I didn't see that manipulating it like I did would crate new solutions. So I have just edited the question. $\endgroup$ – Ury May 22 '16 at 6:38
  • $\begingroup$ but $1-3 \neq \sqrt{1+3}$ isn't it? $\endgroup$ – Siong Thye Goh May 22 '16 at 6:57
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    $\begingroup$ @SiongthyeGoh The question has changed since I put up that comment. Clear questions that change when people put up answers are tiresome. It would be better if the OP put up a new separate question. $\endgroup$ – almagest May 22 '16 at 7:01
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I hope I haven't made a mistake somewhere on the way, but here it goes for stronger condition:

Every two consecutive numbers must add up to a square, so let's call $b_n^2=a_n+a_{n+1}$. Then we get $a_4=a_3+b_1$, $a_5=a_4+b_2$, $a_{n+1}=a_n+b_{n-2}$, so $$a_m=a_3+b_1+b_2+\ldots+b_{m-3}$$ $$b_{n+3}^2-b_{n+2}^2=a_{n+4}-a_{n+2}=b_{n+1}+b_n$$ Now let's explore the sequence $(b_n)$ a bit more: $$b_{n+3}^2-b_{n+2}^2=b_{n+1}+b_n$$ $$b_{n+1}+b_n=(b_{n+3}+b_{n+2})(b_{n+3}-b_{n+2})=(b_{n+5}+b_{n+4})(b_{n+5}-b_{n+4})(b_{n+3}-b_{n+2})$$ and so $$b_2+b_1=(b_4-b_3)(b_6-b_5)(b_8-b_7)\ldots(b_{2n}-b_{2n-1})(b_{2n}+b_{2n-1})$$ $$b_3+b_2=(b_5-b_4)(b_7-b_6)(b_9-b_8)\ldots(b_{2n+1}-b_{2n})(b_{2n+1}+b_{2n})$$ for an arbitrarily large $n$.

In order for that to work, all but finitely many of $b_{n+1}-b_n$ should be $1$, so for all $n\geq N$ it holds: $b_{n+1}=b_n+1$. Let $c=b_N$, then $b_{N+m}=c+m$. So for all m $$(c+m+3)^2-(c+m+2)^2=(c+m+1)+(c+m)$$ $$2c+2m+5=2c+2m+1$$ The contradiction comes from the assumption that the sequence $(a_n)$ exists, so it doesn't.

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  • $\begingroup$ +1 Looks good. At the moment I cannot see any solutions with period $>2$. $\endgroup$ – almagest May 22 '16 at 13:25
  • $\begingroup$ That was really interesting. Thanks. $\endgroup$ – Ury May 22 '16 at 16:22

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