\(A=3n\left(n^2+1\right)-5\left(n^2+1\right)=\left(3n-5\right)\left(n^2+1\right)\)
Để A nguyên tố thì
\(\left[{}\begin{matrix}3n-5=1\\n^2+1=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3n-6=0\\n^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=2\\n=0\end{matrix}\right.\)