Solution a1145e83-47c8-4c93-8820-a1f346e0101d

Points: 1 pt

By definition
\[
\begin{aligned}
b &= ak \\
c &= a\ell
\end{aligned}
\]
for some integers $k,\ell$. Substituting,
\[
b\pm c=ak\pm a\ell=a(k\pm\ell).
\]
We see $k\pm\ell$ exists.

Final answer:
a\mid(b\pm c)