a.
\(n^4+4=n^4+4n^2+4-4n^2\)
\(=\left(n^2+2\right)-\left(2n\right)^2=\left(n^2+2n+2\right)\left(n^2-2n+2\right)\)
Mà n4 + 4 là số nguyên tố
Lại có: \(n^2-2n+2< n^2+2n+2\)
\(\Rightarrow n^2-2n+2=1\)
\(\Rightarrow\left(n-1\right)^2=0\Leftrightarrow n=1\)
b.
\(n=0\Leftrightarrow n^{1994}+n^{1993}+1=1\) (loại)
\(n=1\Leftrightarrow n^{1994}+n^{1993}+1=1+1+1=3\) (thoả mãn)
\(n>1\Rightarrow\left(n^{1994}+n^{1993}+1\right)⋮\left(n^2+n+1\right)\ge7\)
Vậy n = 1