Subdivisions

So a Gaussian integer is simply a complex number with real and imaginary parts that are both integers, i.e. \(\LARGE{{\mathbb Z}[i]=\left\{ a + bi\ |\ a,b \in {\mathbb Z}\right\}}\). It has a “norm”, which is the square of its absolute value, so \(\LARGE{\left|a+bi\right|^2=a^2+b^2}\). The “units” of Gaussian integers are \(\LARGE{1, -1, i, -i}\). Nothing new.

I’m trying to understand Gaussian primes, and am finding a dearth of information out there. I find that, for instance, Gaussian primes include the Gaussian units multiplied by primes if and only if it is congruent to \(\LARGE{3\mod4}\). Why? This is not explained. Here’s my investigation into this and other questions.

It seems fairly intuitive that a Gaussian prime can be a prime number multiplied by a Gaussian unit, e.g. \(\LARGE{(2,3,5,7,11…)\times(\pm1,\pm i)}\). But \(\LARGE{2}\) (and, by extension, \(\LARGE{-2}\), \(\LARGE{2i}\), and \(\LARGE{-2i}\), are not. What’s the deal?

Well, a prime cannot be factored in the domain of its definition. Since we’re dealing with Gaussian integers, \(\LARGE{2}\) must have a factorization within those Gaussian integers. What could that  be?Turns out, \(\LARGE{(1+i)(1-i)=1^2+i-i-i^2=1+1=2}\). That’s of course a special case.

What about 5, 13, 17, 29, etc.? There must be factorizations within Gaussian integers for these. Because we’re dealing with complex integers, if you multiply conjugates, you can add a 1 pretty easily. 5 is 4+1, or \(\LARGE{2^2-i^2}\), therefore \(\LARGE{(2+i)(2-i)=2^2+i-i-i^2=4-(-1)=5}\). We have to get slightly more creative with 13: \(\LARGE{(3+2i)(3-2i)=3^2+2i-2i-(2i)^2=9-(-4)=13}\). \(\LARGE{17=4^2+1}\), which, as mentioned, is simple. \(\LARGE{29=5^2+4}\). 37 is easy. 41? Trickier, but we’re still talking sums of squares: 25+16=41. 53=49+4; 61=25+36; 73=9+64; 89=25+64; 97=16+81. Turns out, you can do this for any prime that is congruent to \(\LARGE{1\mod4}\). Fermat even had one of his famous unproven theorems stating so. It was proven as early as 1747. I won’t pretend to understand those proofs yet, but I will accept that this is true. The one that I understand best is the one explained in this video, and was originally stated in a single sentence by Zagier.

But wait! Our factorizations aren’t unique! What if we went with \(\LARGE{(1-2i)(1+2i)=1^2-2i+2i-(2i)^2=1+4=5}\)? This factorization is the same as multiplying each term in the original factorization shown by \(\LARGE{i}\), which, as previously discussed, is a unit. Likewise, one can factor \(\LARGE{6}\) into \(\LARGE{2}\) and \(\LARGE{3}\) or into \(\LARGE{-2}\) and \(\LARGE{-3}\), and these aren’t (typically) considered unique factorizations.

A way to deal with this is to “normalize” one’s choices. The two suggested methods I’ve found are to choose \(\LARGE{p=a+bi}\) such that \(\LARGE{a}\) is always odd and positive and \(\LARGE{b}\) is always even, or to choose \(\LARGE{p}\) such that \(\LARGE{p\mod(2+2i)=i}\).

Even so, coming up with factorizations is cumbersome and perhaps the simplest thing to do is multiply the integers together and find the gaps to get the primes; an analog of the Sieve of Eratosthenes for Gaussian integers. Thus, we have the following:

Units:

  • \(\LARGE{\pm1,\pm i}\)

Primes:

  • \(\LARGE{1+i}\)
  • \(\LARGE{1+2i}\)
  • \(\LARGE{1+4i}\)
  • \(\LARGE{2+i}\)
  • \(\LARGE{2+3i}\)
  • \(\LARGE{2+5i}\)
  • \(\LARGE{3}\)
  • \(\LARGE{3+2i}\)

All the results from multiplying the units by the primes can also be considered prime.

Because of the above, a Gaussian prime \(\LARGE{a+bi}\) fits one of the following scenarios:

  • \(\LARGE{a=0}\), and \(\LARGE{b\equiv3\mod4}\)
  • \(\LARGE{b=0}\), and \(\LARGE{a\equiv3\mod4}\)
  • \(\LARGE{a^2+b^2}\) is prime

See this video for more. I hope he puts together one on Fermat’s theorem on sums of two squares at some point.

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