I posted this question on MSE, and failed to get the type of answer I wanted. That's why I would like to post it here and wait for the experts to reply. Here's the link to the MSE post, which I anyways copy paste here as well-

I got this interesting arrangement of numbers from $1$ to $32$ in a group in Facebook-

This being an interesting property to look at, I was trying to figure out whether $32$ is something special, or does this hold for other numbers as well. So, let's say, we want to construct one such circle for a general $n$. My idea was to take any $m\in \mathbb N_n=\{1,2,\dots ,n\}$ and find two different integers $x,y\in\mathbb N_n$ so that both $(m+x)$ and $(m+y)$ are perfect squares. Then, we need to find $x^\prime, y^\prime\in \mathbb N_n$ different from $x$, $y$ and $m$ such that both $(x+x^\prime)$ and $(y+y^\prime)$ are perfect squares. Then, we can continue in this manner.

Now, for any two $a,b\in \mathbb N_n$, we have $2\leq a+b\leq 2n$. So, for our given $m$, when we are looking for the mentioned $x$ and $y$, we only need to check through all the perfect squares in the interval $[m,2n]$. So, to reduce our work, we can take our initial choice $m$ to be equal to $n$.

But, now comes the main problem. Let's say, we want to work it out for $\mathbb N_{33}$. So, let's say, our initial $m$ is $33$. The values of $x$ and $y$ can be from the set $\{3,16,31\}$. The question is, which two of these three to choose so that eventually we don't run into a repitition. Note that, in the $\mathbb N_{32}$ case (which is the one in the diagram), taking $m=32$ leaves us with an advantage of having only two choices $4$ and $17$. But, in the next step, for finding the appropriate $x^\prime$ for $x=4$, we have the choices $\{5,12,21\}$ (since $32$ is already taken). The given diagram uses $x^\prime =21$. I tried using $x^\prime =5$ and then using values of my choice, but soon ran into an unavoidable repitition.

So, is there a way to make more circles with other values of $n$, or does $32$ have some profound property which makes it the only possible choice for $n$? I thought, maybe $32$ being a power of $2$ has to do something with it. So, I tried using $n=2,4,8,16$. But, these are trivially NOT solutions since there aren't enough perfect squares in $[n,2n]$ for these values of $n$. Note that this "not enough squares" angle immediately gives a lower bound of $n\geq 19$ by trial and error. As Peter Taylor pointed out, this "not enough squares" argument also rules out $n\le 30$ since $18$ is always a vertex of degree $1$ otherwise. So, we have a lower bound of $n\ge 31$. Also, I was too lazy to even attempt $n=64$.

So, is this case a rare coincidence, or can we have other values of $n$ satisfying this property as well? If there are other values, what family do they belong to? Also, for any case, is that arrangement unique?

Edit:I got one interesting answer that uses graph theory. But, that has a computer program to check that it holds for all $100\geq n\geq 32$. However, I want more "mathy" arguments. I want to see why this happens instead of to only confirm that it happens. I want to know whether it happens for all $n$ or whether there are cases where it breaks down. Please consider these questions as well.

One answer I got is this which (as I already mentioned) is only good enough to check using a computer program that our desired property holds for $100\geq n\geq 32$. But, it does provide a very nice insight into the question using Graph Theory in terms of Hamiltonian cycles. So, added to the questions I asked in the "Edit" paragraph of the main post, one more question I would like to ask on this platform is, **has there been any formal research on this topic?** That is, are there any papers dealing with finding Hamiltonian cycles in this particular case? If yes, please let me know.

pathin $X_n$ starting and ending at distinct neighbours of $n+1$. So I asked the computer about Hamilton paths in $X_n$. For $n$ around $30$-$40$ there are quite a few pairs of vertices that are not connected. Strangely though, for $n$ at least 51 up to 70, there is auniquepair of vertices (always 14 and 31) that are not connected by a Hamilton path in $X_n$. So $X_n$ is very close to being Hamilton connected. $\endgroup$12more comments