TITLE: three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third
QUESTION [5 upvotes]: I need to determine three distinct positive integers $a, b, c$ such
that the sum of any two is divisible by the third.
I tried like with out loss of generality let $a<b<c$
As, $a\mid (b+c)$ so $b+c=ak_1$ for some $k_1\in\mathbb{N}$ similarly $$a+b=k_2 c$$ $$a+c=k_3b$$ so adding them I get $k_1+k_2+k_3=2$, could anyone help me to proceed?

REPLY [3 votes]: $$ak_1-b-c=0 \ \ \ \ \ (1)$$
$$a-bk_2+c=0 \ \ \ \ (2)$$
$$a+b-ck_3=0 \ \ \ \ (3)$$
Using this, for non-trivial solution,
$$\det\begin{pmatrix} k_1 & -1 & -1 \\ 1 & -k_2 &1 \\ 1&1&-k_3\end{pmatrix}=0$$
$$\implies k_1(k_2k_3-1)-1(k_3+1)+1(1-k_2)=0$$
$$\implies k_1+k_2+k_3=k_1\cdot k_2\cdot k_3$$
From the answer by BrianM.Scott of this, we can take $k_1=1,k_2=2,k_3=3$ for any  base $b$ where $b$ is a natural number $>1$
Solving $(1),(2),(3)$ we get $b=2c,a=3c\implies (a,b,c)=(3c,2c,c)$