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Now duplicate of Magic $\mathbb{Z}\times\mathbb{Z}$-square where it has an answer.

Is there an injective map $j:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ satisfying: For every $z\in \mathbb{Z}$ we have $$\lim_{N\to \infty}\sum_{k=-N}^Nj(k,z) = 0 = \lim_{N\to \infty}\sum_{k=-N}^Nj(z,k)\text{ ?}$$

That is, for every $z\in \mathbb{Z}$ there is $N_0=N_0(z)\in \mathbb{N}$, such that for every integer $N\geq N_0$ we have $$\sum_{k=-N}^Nj(k,z) = 0 = \sum_{k=-N}^Nj(z,k).$$

Note. Martin Sleziak has answered this question in another thread: https://mathoverflow.net/a/344049/8628 (no answers could be posted here since this question is on hold).

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    $\begingroup$ I am not sure why this problem was down-voted or why it was put on hold. There is at least one paper in the literature on infinite magic squares (different from this generalization): Antoine Salomon, "Infinite sized magic square," the American Mathematical Monthly (January 2014). hal.archives-ouvertes.fr/hal-01401844v4/document There is also a discussion elsewhere on this site about infinite magic squares. mathoverflow.net/questions/53420/… If this problem is somehow trivial, then someone can provide the solution. $\endgroup$
    – Tri
    Commented Oct 15, 2019 at 15:06
  • $\begingroup$ Thanks @Tri! Seeing the downvotes, I assumed there must be a trivial answer to my question - but please can anyone give a hint to a trivial example or a trivial reason why such a $\mathbb{Z}\times\mathbb{Z}$-square can not exist? $\endgroup$
    – Dominic van der Zypen
    Commented Oct 16, 2019 at 6:52
  • $\begingroup$ One problem is that these limits are never defined, since the summands are distinct integers. $\endgroup$
    – Ben Barber
    Commented Oct 16, 2019 at 10:17
  • $\begingroup$ @BenBarber If $s: \mathbb{N}\to\mathbb{Z}$ is a sequence of integers, then $\lim_{n\to\infty} s(n) = 0$ iff there is $K\in\mathbb{N}$ such that $s(n) = 0$ for all $n\geq K$, or am I mistaken? - At all events, I have provided an alternative formulation of the question that tries to avoid the problem that Ben Barber mentions. Are there any other problems / ambiguities in the question? $\endgroup$
    – Dominic van der Zypen
    Commented Oct 16, 2019 at 20:00
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    $\begingroup$ @DominicvanderZypen I've copied the question here to the other thread. I suggest you delete this one since it doesn't need to be in 2 threads. (For memory, the question was initially downvoted and closed because the sum was not specified to be from $-N$ to $+N$, so the convergence was obviously impossible). $\endgroup$
    – YCor
    Commented Oct 17, 2019 at 22:49

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