Ta có:
\(4n+3⋮2n+1\)
\(\Rightarrow\left(4n+2\right)+1⋮2n+1\)
\(\Rightarrow2\left(2n+1\right)+1⋮2n+1\)
\(\Rightarrow1⋮2n+1\)
\(\Rightarrow2n+1\in U\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}2n+1=-1\Rightarrow n=-1\\2n+1=1\Rightarrow n=0\end{matrix}\right.\)
Vậy \(n\in\left\{-1;0\right\}\)
