On 15/08/2013 16:48, Henry Baker wrote:
All perfect squares of Gaussian integers are Pythagorean triples; are all Pythagorean triples squares of Gaussian integers?
If I'm understanding the question right (maybe not) the answer is "yes, of course, up to multiplying everything by an integer". The general Pythagorean triple is k.(m^2-n^2), k.2mn, k.(m^2+n^2) where k can be any integer. If we ignore k, then m^2-n^2 + 2mn.i = (m+in)^2 so we have the square of a Gaussian integer. If we don't ignore k, we can fail to have the square of a Gaussian integer; e.g., take any primitive triple, e.g. m=2,n=1 giving 3,4,5, and then let k=7 giving 21,28,35. 21+28i isn't the square of a Gaussian integer because if it were we'd have 7|w|^2=|z|^2, hence 7=|z|^2/|w|^2, but both |w|^2 and |z|^2 have to have an even number of factors of 7 because 7 = -1 (mod 4). But this sort of thing -- having a common factor with an odd power of a prime that's -1 mod 4 -- is the only way it can happen. -- g