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