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.

1 Answer
May 15, 2017

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.