Answer
$3m + 5$ is a even number if $m$ is even, because $m$ can be write as $2k$, substituting in the formula, we get $2(3k + 2) + 1$ that is an odd integer;
Work Step by Step
If $m$ is an even integer. $m$ can be write as:
$m = 2 k$
(with $k$ being one integer number)
$k \in \mathbb{Z}$
Now we can develop the formula $3m + 5$:
$\begin{split}
3m + 5 & = 3(2k) + 5 \\
& = 6k + 5 \\
& = 6k + 4 + 1 \\
& = 2(3k + 2) + 1 \\
\end{split}$
Since $3k + 2$ is an integer, then $2(3k + 2) + 1$ is an even number, by the definition of even numbers.
