Ta có
\(A=n^4+4=\left(n^4+4n^2+4\right)-4n^2=\left(n^2+2\right)^2-\left(2n\right)^2\)
\(=\left(n^2-2n+2\right)\left(n^2+2n+2\right)\)
A là số nguyên tố
\(\Leftrightarrow\left[{}\begin{matrix}n^2-2n+2=1\\n^2+2n+2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=1\\n=-1\end{matrix}\right.\)
Vì \(n\in N\) nên n=1