Given that *i*^{2} = -1, for how many integers *n* is (*n* + *i*)^{4} an integer?

Expand (*n* + *i*)^{4} to see what it looks like:

(

=

=

*n*+*i*)^{4}=*n*^{4}+ 4*n*^{3}*i*+ 6*n*^{2}*i*^{2}+ 4*n**i*^{3}+*i*^{4}=

*n*^{4}+ 4*n*^{3}*i*– 6*n*^{2}– 4*n**i*+ 1=

*n*^{4}+ 4n(*n*^{2}– 1)*i*– 6*n*^{2}+ 1This is an integer only if the imaginary part is 0, i.e.,

0 = 4

*n*(*n*^{2}– 1) = 4*n*(*n*+ 1)(*n*– 1), thus*n*= 0, 1, -1.So (*n* + *i*)^{4} is an integer for 3 values of *n*.