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
George C.
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#.

Related questions
Impact of this question
1299 views around the world
You can reuse this answer
Creative Commons License