Solution cdd208db-4a42-439d-92bb-4a2ceaddea40

Points: 1 pt

By definition,
\[
  b=ak
\]
and
\[
  b\pm c=a\ell
\]
for some integers $k,\ell$. Subtracting,
\[
  \pm c=a(\ell-k).
\]
Since $\pm(\ell-k)$ is an integer,
\[
  c=a\bigl(\pm(\ell-k)\bigr).
\]

Final answer:
a\mid c