Dividing!

$c \equiv 0 \mod a$ $b \equiv 0 \mod a$ $bx+cy \equiv 0 \cdot x + 0 \cdot y \mod a$ $0 \mod a$ $\boxed{True}$

Given: $a|b$ and $a|c$ ;

We need to show $a|(bx+cy)$ ,where $x,y \in\ Z$

Since $a|b$ and $a|c$ ; then there exits integers $s$ and $t$ such that $b=as$ and $c=at$ .

Then $bx+cy= asx+aty = a(sx+ty)$ ;

Since $s,x,t \ and \ y$ $\in\ Z$

$\implies$ $sx+ty$ is also $\in \ Z$

Hence, there exists an integer multiplied by $a$ and is equal to $bx+cy$ $\implies$

$a|(bx+cy$ ) $\square$

$a|b,a|c$ so $a|b+c$ . Therefore $a|(b+b+\ldots (x \quad times))+(c+c+\ldots(y \quad times)$ ) or $a|bx+cy$ .

If $x$ and $y$ are negative numbers, just interchange the plus sign with the minus sign in the second last step.