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Informal version. For which $n>1$ can the numbers $1,\ldots, n^2$ be arranged in a square form such that the sums of the numbers in the little squares (consisting of $4$ numbers) are all equal?

Formal version. For any positive integer $n$ we let $[n] := \{1,\ldots,n\}$.

Let $n\in\mathbb{N}$ with $n>1$. If $\psi:[n]\times[n]\to [n^2]$ is a bijection and $k,\ell\in[n-1]$, we let the square sum at $k,\ell$ with respect to $\psi$ be the number $$\text{ssq}_\psi(k,\ell) = \psi(k,\ell) + \psi(k+1,\ell) + \psi(k,\ell+1) + \psi(k+1,\ell+1).$$

We say $n$ has a square sum arrangement if there is a bijection $\psi:[n]\times[n]\to [n^2]$ such that for all $k,k',\ell,\ell'\in[n-1]$ we have

$$\text{ssq}_\psi(k,\ell)=\text{ssq}_\psi(k',\ell').$$

Question. Are there infinitely many integers $n>1$ with a square sum arrangement?

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    $\begingroup$ Do you really mean "small (2 × 2) squares" in the informal version? $\endgroup$
    – Seva
    Commented Jul 2, 2019 at 9:20
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    $\begingroup$ @Seva it should depend on whether we arrange numbers in integer points or in boxes. $\endgroup$
    – Fedor Petrov
    Commented Jul 2, 2019 at 10:28
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    $\begingroup$ I mean the informal version as explicitly described in the formal one :-) I hope the formal one is understandable? $\endgroup$
    – Dominic van der Zypen
    Commented Jul 2, 2019 at 10:43
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    $\begingroup$ Can you provide an example of such an arrangement for some non-tiny $n$? $\endgroup$
    – Greg Martin
    Commented Jul 2, 2019 at 16:19
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    $\begingroup$ Most-perfect magic squares are such, and they do exist as soon as $4|n$. See also OEIS A051235. $\endgroup$
    – Max Alekseyev
    Commented Jul 2, 2019 at 20:38

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