12.3 Given $i^{2}=-1$, the number of $n$ that makes $(n+i)^{4}$ an integer is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(31st American High School Mathematics Examination, 1980)
[Solution] Since $i^{2}=-1$, we have
$$
(n+i)^{4}=n^{4}-6 n^{2}+1+i\left(4 n^{3}-4 n\right),
$$

which is a real number if and only if $4 n^{3}-4 n=0$.
Solving this, we get $n=0,1,-1$. Therefore, there are only three values of $n$ that make $(n+i)^{4}$ a real number, and in these cases, $(n+i)^{4}=n^{4}-6 n^{2}+1$ is an integer.
Hence, the answer is $(D)$.