\(A=\left(n-1\right)\left(n^2+2n+3\right)=\left(n-1\right)\left[\left(n+1\right)^2+2\right]\)
\(n\in N\Rightarrow\left(n+1\right)^2+2\ge3\)
A nguyên tố => \(\Rightarrow\left\{{}\begin{matrix}n-1=1\\\left(n+1\right)^2+2\in P\end{matrix}\right.\)
\(n=2\Rightarrow A=1.\left(3^2+2\right)=11\in P\)
vậy n =2