How many distinct complex integers divide 4?

Question

How many distinct complex integers (n + im) divide 4 + 0i, so that

4 + 0i = (n + mi)(k+li), and
all n,m,k,l are integer.

Answer 1

The prime factorization of 4 in the complex integers (Gaussian integers) is

4 = (1 + i)^2 (1 - i)^2

So the divisors will be 1, 1 + i, 1 - i, (1 + i)^2, (1 - i)^2, (1 + i)^2(1 - i), (1 + i)(1 - i)^2, and 4, plus associates.

Simplifying, this gives the divisors 1, 1 + i, 1 - i, 2i, -2i, 2 + 2i, 2 - 2i, and 4.

Taking associates of all these numbers, we get 1, i, -1, -i, 1 + i, 1 - i, -1 - i, -1 + i, 2i, -2i, 2, -2, 2 + 2i, 2 - 2i, -2 - 2i, -2 + 2i, 4, 4i, -4, -4i.

The particular factorizations are are
4 = (1)(4)
4 = (i)(-4i)
4 = (-1)(-4)
4 = (-i)(4i)
4 = (1 + i)(2 - 2i)
4 = (1 - i)(2 + 2i)
4 = (-1 - i)(-2 + 2i)
4 = (-1 + i)(-2 - 2i)
4 = (2)(2)
4 = (-2)(-2)
4 = (2i)(-2i)

This gives twenty distinct complex integers which divide 4.

If you are only interested in finding divisors up to multiplication by a unit, then 1, 1 + i, 2, 2 + 2i, and 4 are all you get. (Multiplying each of these by i, -1, or -i gives the remaining divisors.)

Answer 2

Clearly : i x -4i = 4, so -i x 4i = 4, 4i x -i = 4 and -4i x i = 4
so i,-i,4i and -4i are 4 solutions
We can do the same for 2i x -2i, ...
so 2i, -2i are also solutions
The integer solutions are the usual divisors of 4 : 1,2 and 4 and -1, -2 and -4
There are other solutions as well :
2(1+i)(1-i)=4 because of the algebraic law (a+b)(a-b)=a²-b²
So we have also 1+i,1-i,2+2i,2-2i
further possible extensions are a(b+i)(1-i/b), but then we need
a(b+1/b)=4 and this is not possible for integer a and b > 1.