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…

4+32*i*=(1+*i*)²(1-*i*)²(3-2*i*)(-1+2*i*)

PROOF for primality test:

So, we say that if a²+b²=P, then a+b*i* is prime and we’ll prove this by contradiction.

Assume that a²+b²=P, but that a+b*i* is NOT prime, or that

(a+b*i*)=(m+n*i*)(x+y*i*)

=(mx-ny)+(my+nx)*i*.

This means that

a=mx-ny

and

b=my+nx

so

a²+b²=(mx-ny)²+(my+nx)²

=m²x²-2mnxy+n²y²+m²y²+2mnxy+n²x²

=m²x²+m²y²+n²x²+n²y²

=m²(x²+y²)+n²(x²+y²)

=(m²+n²)(x²+y²)

Which contradicts our assumption that a²+b²=P.

So if a²+b²=P, then a+b*i* is a complex prime.

on December 3, 2009 at 10:03 PM |Carnival of Mathematics #60 « ∑idiot's Blog[…] 3) Riemann’s zeta function, the lead character in his hypothesis, is connected to primes by Euler’s product formula. If you are interested in the distribution of the primes, Matt Springer at Built on Facts has a post about the function Li(x), as part of his running Sunday Function series. If natural number primes aren’t exciting enough for you, Rich Beveridge at Where The Arts Meet The Sciences has a post for you on Gaussian Primes. […]

on February 3, 2011 at 11:54 AM |you know ‘em you love ‘em you can’t live without ‘em « Where the Arts Meet the Sciences[…] written about the complex or Gaussian primes in a previous post, but I realized that I had never put up a picture of the complex primes, which is what motivated me […]