How can we decide whether the word “factor” becomes the legitimate word factor in such a number system? The reason we can do this it because prime numbers can be defined for this new system as well.

Viewing these so-called Gaussian primes is fun. Many authors have created beatiful graphics such as this.

The basic reason this even makes sense is that we can use the Euclidean algorithm here. First, let's use the same definition of norm as we used earlier for the points, so that \(N(x+iy)=x^2+y^2\).

The difference is that instead of saying simply that \(a=bq+r\) for \(r< b\), we will need to compare the norms of \(r\) and \(b\). Namely, you can write two Gaussian integers \(a\) and \(b\) as \(a=bq+r\), where \(0\leq N(r)<N(b)\), and this process ends by well-ordering to define \(gcd(a,b)\). In this case \(\pm 1\) and \(\pm i\) are all possible stopping points if \(a\) and \(b\) don't share a factor.

Further, if \(g\) and \(h\) are “relatively prime” Gaussian integers (\(gcd(g,h)=\pm 1\) or \(\pm i\)), then there are other such integers \(x\) and \(y\) such that \(gx+hy=1\). So we have a Bezout identity as well to play with.

Computing with Gaussian integers this way is possible, but the naive things one might do are tricky to get to work nicely with the not-so-naive things professional number theorists need factorization in general algebraic structures coming from the complex numbers to do.

Crucially, I am skipping whether we actually have unique factorization in \(\mathbb{Z}[i]\).

This allows a quite different approach to the fact primes of the form \(4n+1\) can be written as a sum of squares. We could use complex numbers instead. Unfortunately, it requires us to take an algebraic fact on faith – the fact we proved using geometry.

Still, it's worth looking at.

• Let \(p\equiv 1\text{ mod }(4)\), as we know.
• We already know that \[r = \left(\frac{p-1}{2}\right)!\] is a square root of \(-1\) modulo \(p\).
• But now, instead of doing geometry, let's look at what that means. It means, by definition of \[r^2\equiv -1\text{ mod }(p)\] that \[p\mid r^2+1\; .\]
• Since \(r^2+1\) is \(r^2-i^2\), let's factor: \[r^2+1=(r+i)(r-i)\]
• Clearly \(p\) does not divide either of those as something of the form \(a+bi\).
• So (crucial!), if we assume the FTA still holds for Gaussian integers, then \(p\) factors in \(\mathbb{Z}[i]\), and has a prime divisor \(a+bi\).
• It's not hard to show that then \(a-bi\) also must divide \(p\). We'll skip this.
• Then \[(a+bi)(a-bi)\mid p^2\Rightarrow a^2+b^2 \mid p^2\]
• Further, this is a nontrivial divisor, since \(a+bi\) was a proper divisor of \(p\). So the only possibility is \[a^2+b^2=p\; .\]