Largest set of natural numbers such that when picking any three, one divides the sum of the other two?

Question

What's the largest set of natural numbers such that:

• no number divides any other
• pick any three numbers, one of those three divides the sum of the other two

Found this problem on puzzling.SE (source) and decided to ask here after playing with it a bit.

Edit: I was able to find examples for sets of sizes $\le6$ with a brute force search, such as:

$\{2, 3, 5, 7, 107, 10693\},\{2, 3, 5, 7, 193, 3467\},\{2, 3, 5, 7, 317, 31693\},\dots$
$\{2, 3, 5, 13, 107, 10267\},\{2, 3, 5, 13, 127, 17267\},\{2, 3, 5, 13, 497, 47707\},\dots$
$\{2, 3, 5, 17, 73, 13867\},$ $\{2, 3, 5, 17, 97, 14353\},\{2, 3, 5, 17, 607, 89833\},\dots$
$\{2, 3, 7, 11, 235, 26309\}, \{2, 3, 7, 11, 437, 60295\}, \{2, 3, 7, 11, 697, 78053\}, \dots$

But I was not able to extend any of these so far to seven elements.

Update: Looks like this thread linked in the comments by Arnaud Mortier contains a proof.

Answer 1

Here is a proof that n=6.

I'll just reproduce the statements of the successive lemmas used in the proof.

Let $S$ be a set satisfying the conditions of the problem.

• Lemma 1: There are at most two even numbers in $S$.
• Lemma 2: If $x_i>x_j,x_k$ all lie in $S$, then $x_i|x_j+x_k$ if and only if $x_i=x_j+x_k$.
• Lemma 3: If there are two even numbers in $S$, the larger even number is the maximal element in $S$.
• Lemma 4: If there exists a working set with $n$ elements, there exists a working set with $n$ elements that contains $2$.
• Lemma 5: $\sharp S=7$ implies that $S$ cannot contain $2$.