<s>[INST] three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third 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?
 [/INST] You should get $2(a+b+c)=k_1a+k_2b+k_3c $ - $(*)$
Now if $k_1=k_2=k_3=2$ we get $a=b=c$.
Suppose $k_1\le k_2\le k_3$, then we must have $k_3 \gt 2$ and $k_1 \lt 2$ else the two sides of $(*)$ cannot be equal.
Since $k_1$ is a non-zero positive integer less than $2$ is must be $1$.
So we have $a=b+c$, and $b+(b+c)=k_2c$, and $c+(b+c)=k_3b$ so that $$b=\frac {k_2-1}2c, \text{ } c=\frac {k_3-1}2b$$ whence $$(k_2-1)(k_3-1)=4$$ and we have $k_2=k_3=3$ with $b=c$, which is not allowed, or $k_2=2, \text { } k_3=5$.
In this case $c=2b$ and $a=b+c=3b$ gives the family of solutions suggeted in the comments.
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