Fix a $k \in \mathbb{N}$

  • How do I find all sets of positive integers $a_{1},a_{2},...,a_{k}$ such that the sum of any triplet is divisible by each member of the triplet.

I couldn't see any pattern. But the simplest example which I could see is the set $\{1,2,3\}$. This satisfies the hypothesis. And even sets like $\{2,4,6\}$ does satisfy.

In search of finding a solution I attempted the following: Let $\{a_{1},a_{2},a_{3}\}$ be a triplet. Then by our hypothesis we should have $$a_{1}\mid (a_{1}+a_{2}+a_{3}), \quad a_{2} \mid (a_{1}+a_{2}+a_{3}) \quad a_{3}\mid (a_{1}+a_{2}+a_{3})$$ Using all this I arrive at $$k_{1}a_{1} = k_{2}a_{2} = k_{3}a_{3}$$ But all this doesn't seem to help.

  • $\begingroup$ You write "sets" but don't use set notation in "$a_1,a_2,\ldots,a_k$". Are the $a_i$ meant to be distinct? $\endgroup$ – joriki Jun 2 '16 at 15:06
  • $\begingroup$ @joriki I have basically copied the question from Ivan Niven's Number theory book. And he doesn't use the set notation there. But let's assume $a_{i}$'s are distinct. This is Section 1.3 Exercise 46, from the book :) $\endgroup$ – crskhr Jun 2 '16 at 15:08

If $a_1\lt a_2\lt a_3$, their sum $S$ satisfies $a_3\lt S\lt 3a_3$. Since $a_3\mid S$, this implies $S=2a_3$, thus $a_1+a_2=a_3$ and thus $a_2\gt a_3/2$. The only divisor of $2a_3$ greater than $a_3/2$ and less than $a_3$ is $\frac23a_3$. Thus we have $a_2=2a_1$ and $a_3=3a_1$. This cannot continute to hold if we add a fourth number, so the only sets with this property are those of the form $\{a_1,2a_1,3a_1\}$ with $a_1\in\mathbb N$.


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