 Solution

Let $d=\gcd (c,n)$, $c=d{c}^{\prime }$, and $n=d{n}^{\prime }$. The preceding part gives $\gcd (c^{\prime },n^{\prime })=1$. The integer congruence on the left is equivalent toBy Euclid's lemma, coprimality of $c^{\prime }$ and $n^{\prime }$ makes this equivalent to $n^{\prime }\mid a-b$. The reverse implication is also immediate by multiplying a divisibility relation by $d{c}^{\prime }$. ThereforeThis is cancellation of a common factor in a congruence. The printed equality sign in the right-hand expression is interpreted as equality of residue classes, hence the displayed integer congruence. Ordinary equality of the integers $a,b$ would be false: the following example has $a\ne b$.