I proved that it is possible to express any integer as the sum of the squares of two Gaussian integers (see proof below). My question is whether or not there are other simpler proofs.
My proof:
For any integer n,
$\quad (n+0i)^2 + (0+[n-1]i)^2 =$ $n^2 - (n^2-2n+1) =$ $2n-1$
(1) Therefore, all odd integers can be written as the sum of the squares of two Gaussian integers.
For any integer n,
$\quad ([n+1]+0i)^2 + (0+[n-1]i)^2 =$ $(n^2+2n+1) - (n^2-2n+1) =$ $4n$
(2) Therefore, all integers $\equiv 0\pmod 4$ can be written as the sum of the squares of two Gaussian integers.
For any integer n,
$\quad(n+[n-1]i)^2 + (n-[n-1]i)^2 =$
$\quad(n^2+2n[n-1]i-[n-1]^2) + (n^2-2n[n-1]i-[n-1]^2) =$
$\quad 2(n^2-[n-1]^2) =$
$\quad 2(n^2-[n^2-2n+1]) =$
$\quad 2(2n-1) =$
$\quad 4n-2$
(3) Therefore, all integers $\equiv 2\pmod 4$ can be written as the sum of the squares of two Gaussian integers.
Combining results (1), (2) and (3), we have shown that every integer can be expressed as the sum of the squares of two Gaussian integers.
Q. E. D.