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My question concerns the problem of 'the square-sum problem' which is described in details in this video: https://youtu.be/G1m7goLCJDY On this video is told that a series of sequential numbers from 1 to 15 we can rearrange in such a way that the sum of two neighboring ones will be an perfect square. Here is a solution {9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8}. That is not the only number with this property. The question goes like 'does there exist the biggest N with these properties. If there is not, how to prove it?'. P.s. excuse for my not-inteligence on this comment, i'm not a student, i'm just a scholar who spent his time dealing with math and thus i don't have a perfect english and i may translate some verbs in not appropriate way. I need your help,hope you understand the statement correctly, thanks for your attention!!!!

Artem Vladimirov
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    Dear Artem, welcome to Math.StackExchange! It is customary to explain the problem in full here, without resorting to an external link. You should also comment your previous efforts and relevant thoughts about the problem. – Jose Brox May 27 '19 at 14:18
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    Please include the complete statement of the question in the post, even though Numberphile is a well-established Youtube channel which clip is not likely to be taken down. – Lee David Chung Lin May 27 '19 at 14:18
  • The main question seems to be quite difficult, but here is one property which is easy to prove: on the graph of $N$ vertices relating to this problem, if $N=2n^2$ then the graph has exactly ${n\over3}(n+1)(2n+1)-1-n^2$ edges, and if $N=2n(n+1)$ then the graph has exactly ${n\over3}(n+1)(2n+1)-1$ edges. This probably isn't useful on its own, but it does show that the number of edges (which is monotone increasing with N) is exactly $\Theta(N^{1.5})$ – stanley dodds May 27 '19 at 15:45
  • The exact asymptotic relationship between the number of vertices and edges, $N$ and $E$, is $E\sim{{N^{1.5}}\over{3\sqrt{2}}}$ – stanley dodds May 27 '19 at 15:58
  • @stanleydodds My idea of solving this problem is based on proving that this is Hamilton graph. I am thinking that for a large enough n the degree of each vertex of the graph is bigger than n/2, which is a sufficient criterion – Artem Vladimirov May 27 '19 at 16:03
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    @ArtemVladimirov This would be great if it were true, but unfortunately as I pointed out, with $N$ vertices the number of edges is bounded by $\mathcal{O}(N^{1.5})$. Hence there are at most $\mathcal{O}(\sqrt{N})$ edges connected to the vertex with the smallest degree (by pigeonhole principle). So in fact for large enough $N$, your statement is never true. – stanley dodds May 27 '19 at 16:08
  • Your solution sequence can be extended to $N=17$ by adding $16$ before its first term and $17$ after its last term. – Alex Ravsky Jun 08 '19 at 13:10
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    @AlexRavsky there is a solution for n=2500, this result was gained by supercomputer in University of Twente – Artem Vladimirov Jun 08 '19 at 13:25
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    Does this answer your question? Arranging numbers from $1$ to $n$ such that the sum of every two adjacent numbers is a perfect power – pisoir Jun 07 '20 at 16:11

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