About 10 years ago, I latched onto the concept of Gaussian Primes.

A Complex (or Gaussian) Prime works much the same way that a real-valued prime does. It has no divisors other itself and the unit element.

In the real numbers, the units are 1 and -1. In the Complex number system, the units are 1, -1, *i*, –*i*.

In order to tell if a complex number (a+b*i*) is prime, simply compute a²+b². If the result is a real-valued prime, then the original complex number (a+b*i*) is prime. The proof of this is actually pretty simple and interesting.

Today, in Intermediate Algebra, we were caculating some multiplication problems involving complex numbers.

One of these was (5-3*i*)(1+*i*)=8+2*i*

I noticed that the answer was factorable as 2(4+*i*) and then realized that these were two different factor pairs for the complex number 8+2*i*. That meant that I should be able to turn one factor pair into the other by shifting around the prime factors of the original number.

An example of this is that 24=6*4 and 24=3*8, so I can create one factor pair from the other by shifting a factor of 2.

In considering 6*4, if we look at the 6 as 3*2, then 6*4=(3*2)*4=3*(2*4)=3*8.

Pretty simple, in the real number system.

In the example we looked at in class today we said that

8+2*i*=(5-3*i*)(1+*i*)=2(4+*i*).

Now, I knew that 1+*i* is prime (because 1²+1²=2 which is prime) and I suspected that the 2 would factor into (1+*i*)(1-*i*), which would mean that 5+3*i*=(4+*i*)(1-*i*), which it does.

So the prime factorization of 8+2*i* is (1+*i*)(1-*i*)(4+*i*).

Each of these factors is prime and can be checked using the little trick of a²+b²=P.

Here’s a second problem, I’ll include the factorization below…

Find the prime factorization of 4+32*i* given that:

(-4+8*i*)(3-2*i)=*4+32*i*

I mentioned earlier that the proof for this is simple and interesting. I’ll include it below as well… (more…)