Linear combination of divisible numbers

Statement

Prove that if $a \mid b$ and $a \mid c$, then $a \mid (mb + nc)$ for every choice of integers $m$ and $n$.

Solution

From $a \mid b$ and $a \mid c$ there are integers $x$ and $y$ with $b = ax$ and $c = ay$. Then, for all $m, n \in \mathbb{Z}$, $mb + nc = m(ax) + n(ay) = a\,(mx + ny).$ Since $mx + ny$ is an integer, $a$ divides $mb + nc$. $\blacksquare$

(In particular, with $m = n = 1$, we recover that $a$ divides $b + c$.)