Prove that if an integer of the form \(6k + 5\), then it is necessarily of the form \(3k-1\), but not conversely.
Proof
Suppose \(n = 6k + 5\). Observe that
$$
\begin{align*}
n &= 6k + 5 \\
&= 3(2k + 2) - 1
\end{align*}
$$
If we let \(m = 2k + 2\), then \(n = 3m - 1\) as required. Conversely, a counter example is \(2 = 3(1) - 1\). There isn’t a \(k\) such that \(2 = 6k + 5\) since this would imply that \(6k = -3\) so \(k\) is not an integer. \(\ \blacksquare\)