I came across a question today:
Let $a_{1},a_{2},a_{3}>1$ be three pairwise relatively prime integers.Let $r_{1},r_{2},r_{3}$ be the smallest positive integers such that there exist integers $$s_{ij}\ge 0,1\le i,j\le 3 ,i\not =j$$with
$$r_{1}a_{1}=s_{12}a_{2}+s_{13}a_{3}$$
$$r_{2}a_{2}=s_{21}a_{1}+s_{23}a_{3}$$
$$r_{3}a_{3}=s_{31}a_{1}+s_{32}a_{2}$$ I want to prove if $s_{13}=0$, then $s_{23}=0.$
I have tried to prove by contradiction. Suppose $s_{23}\not =0$, it's easy to see $r_{2}\le s_{12}$.
$if \ s_{21}>0$, then $$(r_{1}-s_{21})a_{1}=(s_{12}-r_{2})a_{2}+s_{23}a_{3}$$ contradict the minimality of $r_{1}$. But I can't prove$s_{21}>0$, How can I do it?