It is well known that a positive rational number is a sum of four squares. The Gaussian integers generalize the rational integers. What can be said about the decomposition of a gaussian integer in sum of squares of gaussian integers?
3
$\begingroup$
$\endgroup$
-
3$\begingroup$ For a start, every square has even imaginary part, so anything with odd imaginary part is not a sum of squares. $\endgroup$ – Chris Eagle Jan 21 '13 at 10:38
-
$\begingroup$ A related question: mathoverflow.net/questions/14456/… $\endgroup$ – KCd Jan 21 '13 at 16:05
2
$\begingroup$
$\endgroup$
See here for a proof that every $a+2bi$ with $a$ odd is a sum of $2$ squares, with $a$ even, a sum of three squares.
Ivan Niven, Integers of quadratic fields as sums of squares, Transactions of the AMS 1940, proves that in an imaginary quadratic field every integer $a+2b\sqrt{-m}$ is a sum of three squares. He starts with the Gaussians. The paper should be freely available at the AMS website.
answered Jan 21 '13 at 11:37
