Question

What possible values can the difference of squares of two Gaussian integers take?

The difference of squares of any two integers can take the form `4n+k` for any integer `n` and `k in {0, 1, 3}`. Specifically, the difference of squares of two integers cannot be of the form `4n+2`.

Is there a simple characterisation of possible differences of squares for Gaussian integers, i.e. complex numbers of the form `m+ni`, where `m, n` are integers ?

Conjecture: Any Gaussian integer of the form `m+2ni` where `m, n` are integers is expressible as the difference of two squares of Gaussian integers.

Answer 1

See explanation...

Explanation:

Here are some possibilities:

• `(n+1)^2-n^2 = 2n+1`

• `(n+1)^2-(n-1)^2 = 4n`

• `((n+1)+ni)^2 - (n+(n+1)i)^2 = 4n+2`

So we can get any (real) integer as a difference of squares of Gaussian integers.

More generally:

`(a+bi)^2-(c+di)^2 = (a^2-b^2-c^2+d^2)+2(ab+cd)i`

Consider the various possible combinations of odd and even `a, b, c, d` and the resulting values of `(a^2-b^2-c^2+d^2)` and `(ab+cd)` modulo `4` and `2` respectively:

`((a_2, b_2, c_2, d_2, (a^2-b^2-c^2+d^2)_4, (ab+cd)_2),(0,0,0,0,0,0),(0,0,0,1,1,0),(0,0,1,0,3,0),(0,0,1,1,0,1),(0,1,0,0,3,0),(0,1,0,1,0,0),(0,1,1,0,2,0),(0,1,1,1,3,1),(1,0,0,0,1,0),(1,0,0,1,2,0),(1,0,1,0,0,0),(1,0,1,1,1,1),(1,1,0,0,0,1),(1,1,0,1,1,1),(1,1,1,0,3,1),(1,1,1,1,0,0))`

So if `(ab+cd)` is odd, then `(a^2-b^2-c^2+d^2) = 0, 1` or `3` modulo `4`.

So the conjecture in the question is false: If the difference of squares of two Gaussian integers has an imaginary part of the form `4k+2` then the real part is of the form `4k+0`, `4k+1` or `4k+3`. Specifically not of the form `4k+2`.