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Let $S \subset \mathbb{Z}$ be a set of integers such that for every pair of distinct elements $a, b \in S$, the sum $a + b$ is a perfect square, and no two such sums are equal.

Does there exist an infinite such set $S$? If not, what is the largest possible size of such a finite set?

What I tried:

  • I first considered small sets like $S = \{a, b, c\}$ and tried to assign integer values such that $a+b, a+c, b+c$ are all distinct perfect squares. I managed to find some triples, but extending this to 4 or more elements became increasingly difficult.
  • I also looked at the sequence of perfect squares and tried to reverse-engineer sets $S$ by treating square numbers as pairwise sums $a + b$, but this often led to inconsistencies or repeated sums.
  • I checked OEIS and known constructions in additive number theory, but couldn’t find this exact configuration.
  • Thought about relaxing the condition to allow equal sums or just require some subset to have the property, but that seemed to miss the spirit of the original question.
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    $\begingroup$ This one has more informative answer. $\endgroup$ Commented Jun 12 at 17:53

1 Answer 1

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No. If $S=\{x_1,x_2,\dots\}$ where $x_k<x_{k+1}$ for all $k\ge 1$, then there exists only a finite number of integers $x$ such that $x+x_1$ and $x+x_2$ are perfect squares, because the difference $x_2-x_1$ cannot be obtained between two perfect squares above $(x_2-x_1)^2$.

Note that in any (finite) set $S$ whose pairwise sums are all perfect squares, all even numbers must have the same residue $r$ modulo $4$, and there cannot be more than one odd number, with residue $r+1$ modulo $4$.

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