Elementary number theory -Divisibility

Question

For arbitrary integers a, b and c, either provide a proof or a counter example.

If a|(b+c), then a|b and a|c

My solution was Let counter example a∤(b+c) then,

b+c ≠ ma (m in Z)
∴ b ≠ ma-c ∴ a∤b

and c ≠ ma-b ∴ a∤c

Is my solution correct

Answer 1

I think your proof is incorrect. Just because we have that $b$ is not in the form $ma-c$ for some $m\in\mathbb{Z}$ doesn't mean that it can't be in the form $ma$ for some $m\in\mathbb{Z}$ which is what is being said by $a\nmid b$. The first thing you should check when asked to prove or disprove is concrete examples. Let $a=2$ and $b=c=1$, then $a\mid (b+c)$ but $a\nmid b$ and $a\nmid c$ making the claim false.

Answer 2

With a counter-example you can just provide particular values for $a,b,c$. For example $(a,b,c) = (3,4,5)$ is a counter-example disproving the assertion, since $a \mid b{+}c$ (that is, $3\mid 9$) but $a\nmid b$ and $a\nmid c$.