will this always happen?

an even integer must be in the form $2k$ where $k\in\mathbb{Z}$

$\therefore$ it is enough to show that $4n(4n+2)=2k$

So, $\begin{aligned} 4n(4n+2)\\& =2\cdot 2n(4n+2) \\& =2(8n^2+4n)\\& =2k \end{aligned}$

$\therefore$ $4n(4n+2)$ will always be a even integer where $n\in\mathbb{Z}$