What is the proof for: $\,a\mid b,c\implies a\mid b\pm c$ In my spare time, I'm working my way a book "mathematical introduction to cryptography" in which the following proposition is given:

If $a\mid b$ and $a\mid c$, then $a\mid (b+c)$ and $a\mid (b-c)$

It is left to the reader to show the proof, however, my maths skills have almost disappeared over the last 15 years since leaving uni.
 A: If $c=ka$ and $b=ja$ then $b+c=\cdots \; ?$, $b-c=\cdots \; ?$
In words, if $a$ is a factor of $c$ and a factor of $b$, the distributive laws mean it is a factor of both $a+b$ and $a-b$.
Example $\;\; 6=\color{red}{2}\times 3, 12=\color{red}{2}\times 6$ so $6+12=\color{red}{2}\times(3+6)$ and $6-12=\color{red}{2}\times(3-6)$
A: Hint $\rm\,\ a\mid b,c\ \Rightarrow\ \dfrac{b}a,\,\dfrac{c}a\in\Bbb Z\ \Rightarrow\ \dfrac{b\pm c}{a}\, =\, \dfrac{b}a \pm \dfrac{c}a\in \Bbb Z\ \Rightarrow\ a\mid b\pm c$
Remark $\ $ So we see that this divisibility law is a consequence of $\rm\,\Bbb Z\,$ being closed under addition and subtraction.
A: It would be obvious if you translated the problem into the language of modular arithmetic rather than that of divisibility.
Alternatively, it would be useful to introduce new variables: e.g. let $d$ be the* integer such that $b = ad$, and so forth. then work from there. 
*: In the special case that $a = 0$, there isn't a unique choice for $d$. But we can treat this case specially. Or just let $d$ be any integer with that property.
A: 
If $a\mid b$ and $a\mid c$ then $a\mid b\pm c$

Show: $$a\mid b \Longrightarrow\exists k\in\mathbb{N}\;\text{such that}\;b=a\cdot k$$$$a\mid c \Longrightarrow\exists k'\in\mathbb{N}\;\text{such that}\;c=a\cdot k'$$$$b\pm c=a\cdot k\pm a\cdot k'=a(k\pm k')\Longrightarrow a\mid b\pm c\;\;\;\;\;\Box$$Remember that in case $b-c$; $b>c$
