Answer
$4(n^2 + n + 1) - 3n^2 = (n + 2)^2 $
$n$ is a integer, so $(n+2)^2$ and $4(n^2 + n + 1) - 3n^2$ are both perfect squares.
Work Step by Step
$4(n^2 + n + 1) - (3n^2)$ is a perfect square if it's equal to one integer squared
Let $n$ be any integer
$\begin{split}
4(n^2 + n + 1) - 3n^2 & = 4n^2 + 4n + 4 - 3n^2 \\
& = n^2 + 4n + 4 \\
& = (n)^2 + 2(n.2) + (2)^2 \\
\end{split}$
$4(n^2 + n + 1) - (3n^2)$ is a perfect square if it's equal to one integer squared
$(n + 2)^2$ is a perfect square, then $4(n^2 + n + 1) - (3n^2)$ is a perfect square too.
$4(n^2 + n + 1) - 3n^2 = (n + 2)^2 $
