Prove divisibility: if $a\mid (b-d)$ and $a\mid (c-e)$, then $a\mid (bc-de)$

Question

I have this math question. It states:

Show that for any $a , b ,c, d, e \in \mathbb{Z^+}$, if $a\mid (b-d)$ and $a\mid (c-e)$, then $a\mid (bc-de)$.

I'm not 100% sure as to how to start this problem. At first I tried to rewrite $a\mid (b-d)$ to $b-d = a\cdot x$ and $a\mid (c-e)$ to $c-e = a\cdot y$. Then I multiplied them and got $a^2xy=(b-d)(c-e)$. I'm not sure if this helps though. Thanks.

Answer 1

Take $k,l$ such that $ak=b-d$ and $al=c-e$. Then $$akc=bc-dc\tag{1}$$ and $$ald=cd-ed\tag{2}$$ Summing (1) and (2), we obtain $a(kc+ld)=bc-ed$, so $a|(bc-ed)$.

Answer 2

Taking cue from you results and details, we can say
$b-d = ax => b=d+ax$
$c-e = ay => c=e+ay$
$bc-de=de + ady + axe + a^2xy - de = ady + axe + a^2xy = a(dy + xe + axy)$
This implies that $a|(bc-de)$

Answer 3

$bc-de=(b-d)c+d(c-e)$. Both summands on the right side are divisible by $a$; therefore so is their sum.

Answer 4

$b\equiv d\mod a$ and $c \equiv e \mod a$ imply $bc\equiv de \mod a$.

Answer 5

Following what you started, we have $b-d=ax$ and $c-e=ay$. Now we write

\begin{eqnarray*} bc-ed &=& bc - be + be - ed \\ &=& b(c-e) + e(b-d) \\ &=& bay + eax \\ &=& a(by+ex), \\ \end{eqnarray*}

so $a$ divides $bc-ed$.