Mystery Math: Gaussian Integers

GAUSSIAN INTEGERS

Gaussian integers can be visualized in the complex plane, with their real components on the horizontal axis, and their imaginary components orthogonal along the vertical axis.

[This post is targeted at a level 3 student who has some familiarity with complex numbers.]

Gaussian integers are complex numbers of the form $a+bi,$ where $a$ and $b$ are integers. For example, $2-3i, 4+5i, \ 17,\ 0$ are all Gaussian integers, while $\frac{4}{3}$ , $\sqrt{2}$ , and $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ are not. One can add, subtract and multiply Gaussian integers just like all other complex numbers. For example:
Addition: $(2-3i)+(4+5i)=6+2i$ ,
Subtraction: $(2-3i)-(4+5i)=-2-8i$ ,
Multiplication: $(2-3i)\cdot(4+5i) = 8+ 10i-12i-15i^2=23-2i$ .

As you can see, the result will again be a Gaussian integer. However, if you try to divide two Gaussian integers, the result will not always be a Gaussian integer:
Division: $\frac{2-3i}{4+5i}=\frac{(2-3i)(4-5i)}{(4+5i)(4-5i)}=\frac{-7-22i}{16+25}=-\frac{7}{41}-\frac{22}{41}i$ .

Like all complex numbers, Gaussian integers have the following properties:

1. The conjugate of $a+bi$ is $\overline{a+bi}=a-bi$ , which is again a Gaussian integer.
2. The norm of $a+bi$ is $N(a+bi)=(a+bi)(a-bi)=a^2+b^2$ , which is a non-negative integer.
3. The absolute value of a Gaussian integer is the (positive) square root of its norm: $|a+bi|=\sqrt{a^2+b^2}$ .
4. There are no positive or negative Gaussian integers, and one cannot say that one is less than another. One can, however, compare their norms.

We say that a Gaussian integer $x$ is a unit, if $\frac{1}{x}$ is also a Gaussian integer. The only units are $1,\ -1,\ i,\ -i$ .

A Gaussian integer $(a+bi)$ is a multiple of a Gaussian integer $(c+di)$ if $(a+bi)=(c+di)\cdot (e+fi)$ for some Gaussian integer $e+fi$ . In this case we say that $c+di$ divides $a+bi$ , and use the notation $(c+di) \mid (a+bi)$ .

A Gaussian integer is called prime, if it is not equal to a product of two non-unit Gaussian integers. It is called composite otherwise. Clearly, multiplying by a unit does not change the primality. Note that the same definition for the usual integers implies that $-5$ is a prime integer. This may seem a bit strange, but no other definition makes sense for the Gaussian integers. (Keep in mind that there is no such thing as a positive Gaussian integer!) Note that a number may be prime as a usual integer, but composite as a Gaussian integer, for example $5=(2+i)(2-i)$ .

There are three kinds of Gaussian primes:
1) $1+i, \ 1-i, \ -1+i,\ -1-i$ . They are all the same up to multiplication by a unit, so we can say $e\cdot (1+i),$ where $e$ is a unit.
2) $e\cdot p$ , where $p$ is a usual prime ( $p\in {\mathbb N}$ ), and $p=4k+3$ .
3) $e\cdot(u+vi)$ or $e\cdot(u-vi)$ , where $u$ and $v$ are natural numbers such that $u^2+v^2=p,$ for a prime $p\in{\mathbb N}$ , $p=4k+1$ . Such $u$ and $v$ exist, and are unique, up to switching $u$ and $v$ , for every prime $p=4k+1$ . This classical result is called the Fermat Two Squares Theorem. It was noticed and announced by Pierre Fermat in 1640 and first proven by Leonhard Euler in 1747.

For those who had learned some abstract algebra, the algebraic properties of Gaussian integers (usually denoted by ${\mathbb Z}[i]$ ) make it a commutative ring, moreover, a domain. Furthermore, just like the usual integers, all Gaussian integers can be decomposed into a product of Gaussian primes, uniquely up to units. The formal way of saying this is that ${\mathbb Z}[i]$ is a unique factorization domain (UFD for brevity).

The classification of Gaussian primes is far from obvious, and so is the unique factorization property. (Speaking of which, do you know how to prove it for the usual integers? You know it to be true from experience, but it is actually not easy to prove!) These theorems will be proven in the upcoming post.

Worked Examples

1. Suppose $n\neq 0$ is a usual integer. Show that a Gaussian integer $a+bi$ is a multiple of $n$ if and only if both $a$ and $b$ are multiples of $n$ .

Solution: By definition, one Gaussian integer is a multiple of the other if and only if their ratio is also a Gaussian integer. Observe that $\frac{a+bi}{n}=\frac{a}{n}+\frac{b}{n}i$ , so $\frac {a}{n}$ and $\frac {b}{n}$ are integers, which means that both $a$ and $b$ are multiples of $n$ .

2. Suppose $a+bi$ is a multiple of $c+di$ . Show that $N(a+bi)$ is a multiple of $N(c+di)$ .

Solution: If $(a+bi)=(c+di)(e+fi)$ , then $(a-bi)=(c-di)(e-fi)$ . Therefore,
$N(a+bi)=(a+bi)(a-bi)=(c+di)(e+fi)\cdot (c-di)(e-fi)=$
$=(c+di)(c-di)\cdot (e+fi)(e-fi) = N(c+di)N(e+fi).$
Because $e$ and $f$ are integers, $N(e+fi)=e^2+f^2$ is an integer.

This is the multiplicative property of the norm, which was mentioned in Complex Numbers Test Yourself 2.

3. Decompose $4+3i$ into a product of primes.

Solution: Because $N(4+3i)=25,$ the norm of any prime that divides $4+3i$ must divide $25$ . Looking at the classification of primes, because $5$ is $1$ modulo $4$ and $5=2^2+1^2$ , there are two such primes, up to units: $2+i$ and $2-i$ . Dividing $4+3i$ by $2+i$ , we get $\frac{11}{5}+\frac{2}{5}i,$ which is not a Gaussian integer. Dividing $4+3i$ by $2-i$ , we get $1+2i=i(2-i)$ . So $4+3i=i(2-i)^2.$

4. Suppose a Gaussian integer $x$ divides a Gaussian integer $y$ . Show that $\bar{x}$ divides $\bar{y}.$

Solution: From Complex Numbers – Worked Example 3, we know that conjugate distributes across multiplication, hence if $y= xz$ , then $\bar{y}=\bar{x}\bar{z}$ .