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How many different natural numbers are there such that the difference of any two is a perfect square?

I could find that with 'sums' instead of 'differences' this has been asked by Erdos and L. Moser (see Guy: Unsolved Problems in Number Theory, 3rd ed., p. 268), and that is open, with six being the current record.

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    $\begingroup$ Just one unless you consider only the positive differences. $\endgroup$ Commented Oct 19, 2020 at 20:23
  • $\begingroup$ Not quite the same thing, but there's a recent paper considering a mod p analogue of this problem (originally considered by Lydia Goncharova in unpublished work) Kiritchenko, Valentina; Tsfasman, Michael; Vlăduţ, Serge; Zakharevich, Ilya Quadratic residue patterns, algebraic curves and a K3 surface. Finite Fields Appl. 101 (2025), Paper No. 102517, 29 pp. doi.org/10.1016/j.ffa.2024.102517 arxiv.org/abs/2403.16326, where it gives an exact count of the number of solutions for n=4. It notes that the solutions are rational points on a K3 surface, which is still true over Z. $\endgroup$ Commented Sep 24, 2025 at 14:18
  • $\begingroup$ I would assume but have not checked that for n > 4 standard heuristics will suggest that there should be only finitely many solutions. $\endgroup$ Commented Sep 24, 2025 at 14:19
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    $\begingroup$ For more on the question of finding all four-tuples of integers such that the difference of any two is a square (which is equivalent to finding rational points on a K3 surface) see arxiv.org/pdf/math/0606700. There are infinitely many such but any sort of classification is out of the reach of current techniques. $\endgroup$ Commented Sep 24, 2025 at 21:30

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Four is possible, an example is $0,451584,462400,485809$. This solution comes from the example of a cuboid with integer sides, space diagonal and two out of three face diagonals given here. I don't know for $5$ or more but it is probably a very hard question.

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  • $\begingroup$ Zero is not natural. ####### But for arbitrary natural $\ t\ $ we may have: $ \{t\quad t\!+\!451584\quad t\!+\!462400\quad t\!+\!485809\}$. $\endgroup$ Commented Sep 24, 2025 at 17:14
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    $\begingroup$ @WlodAA: it depends - see math.stackexchange.com/q/283 $\endgroup$ Commented Sep 24, 2025 at 18:50
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    $\begingroup$ @MaxAlekseyev, either way, I showed how to disregard that notation nuance (nuisance!) about natural numbers, positive integers, and non-negative -- at least on this occasion we can forget about it. :) $\endgroup$ Commented Sep 24, 2025 at 19:46
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    $\begingroup$ If you subtract these numbers from 485809, you get Euler's solution, linked to by Alison in her third comment to my question. $\endgroup$ Commented Sep 25, 2025 at 6:47
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Minimum largest of a set of $n$ nonnegative integers such that the difference of any pair is a square (please, note that here I also include the difference of any element with itself), together with the optimum set satisfying the above.
$n=1 \to \{0\}$,
$n=2 \to \{0, 1\}$,
$n=3 \to \{0, 9, 25\}$,
$n=4 \to \{0, 23409, 34225, 485809\}$ (or even Antoine Labelle's solution $\{0, 451584, 462400, 485809\}$).
If the fifth term of the sequence $[0,1,25,485809,\ldots]$ exists, it must be above $9 \cdot 10^{10}$.

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