Exercise 1.2.21 (Integers of the form $6k+ 5$)

Proof. If $n=6k+5$ for some integer $k$, then $n=3(2k+2)-1=3K-1$, where $K=2k+2$, so $n$ is necessarily of the form $3k-1$.

The converse is false. As a counterexample, $8=3\cdot 3-1$ is of the form $3k-1$, but is of the form $6k+2$ (for $k=1$), so $8$ is not of the form $6k+5$ by unicity of the remainder in the division algorithm. □