What are Gaussian Integers?

A Gaussian integer is a complex number whose real and imaginary parts are both integers. We denote the set of Gaussian integers as $\mathbb{Z}[i]$. A Gaussian integer $z$ is a number of the form $z = a + bi$ where $a, b \in \mathbb{Z}$.

Let's consider two Gaussian integers $\alpha = 3 + 4i$ and $\beta = 1 - 2i$.

(a) Find the product $\gamma = \alpha\beta$

To find the product, we multiply the complex numbers:

$$\gamma = (3 + 4i)(1 - 2i) = 3(1) + 3(-2i) + 4i(1) + 4i(-2i)$$

$$= 3 - 6i + 4i - 8i^2 = 3 - 2i + 8 = 11 - 2i$$

So, $\gamma = 11 - 2i$. Since both the real part (11) and the imaginary part (-2) are integers, $\gamma$ is a Gaussian integer.

(b) Is a quotient of Gaussian integers a Gaussian integer?

Given $\alpha = 3+4i$ and another Gaussian integer $\gamma = 11+2i$, let's determine if their quotient $\frac{\gamma}{\alpha}$ is a Gaussian integer.

$$\frac{11+2i}{3+4i} = \frac{11+2i}{3+4i} \times \frac{3-4i}{3-4i} = \frac{33 - 44i + 6i - 8i^2}{3^2 + 4^2}$$

$$= \frac{33 - 38i + 8}{25} = \frac{41 - 38i}{25} = \frac{41}{25} - \frac{38}{25}i$$

The real part is $\frac{41}{25}$ and the imaginary part is $-\frac{38}{25}$. Since these are not integers, the quotient is not a Gaussian integer.

The Norm of a Gaussian Integer

The norm of a Gaussian integer $z = a + bi$ is defined as $N(z) = a^2 + b^2$. It is the square of the modulus of the complex number, i.e., $N(z) = |z|^2$.

(c) Plotting Gaussian Integers with Norm < 3

We need to find all integers $a, b$ such that $a^2 + b^2 < 3$.

If $a=0$, $b^2 < 3 \implies b \in \{-1, 0, 1\}$. Points: $0, \pm i$.
If $a=1$, $1+b^2 < 3 \implies b^2 < 2 \implies b \in \{-1, 0, 1\}$. Points: $1, 1\pm i$.
If $a=-1$, $1+b^2 < 3 \implies b^2 < 2 \implies b \in \{-1, 0, 1\}$. Points: $-1, -1\pm i$.
If $|a| \ge 2$, $a^2 \ge 4$, so no solutions.

The Gaussian integers with norm less than 3 are: $0, \pm 1, \pm i, 1\pm i, -1\pm i$. There are 9 such points.

(d) Show that $N(a+bi)$ is an integer

For any Gaussian integer $z=a+bi$, the norm is $N(z) = a^2+b^2$. Since $a,b \in \mathbb{Z}$, their squares $a^2$ and $b^2$ are also integers. The sum of two integers is an integer, so $N(z)$ is always an integer.

(e) Show that a number of the form $n = c^2+d^2$ is not a Gaussian prime

A Gaussian integer is prime if it is not a unit ($\pm 1, \pm i$) and its only divisors are units and associates (the number itself multiplied by a unit). An integer $n$ can be factored in $\mathbb{Z}[i]$ as $n = (c+di)(c-di)$. Since $c,d$ are integers, $c+di$ and $c-di$ are Gaussian integers. If $n$ were a Gaussian prime, one of these factors would have to be a unit. $N(c+di) = c^2+d^2=n$ and $N(c-di)=c^2+d^2=n$. For the factors to be units, their norm must be 1. But $n=c^2+d^2$ is not necessarily 1. For example, $5 = 2^2+1^2 = (2+i)(2-i)$. Neither factor is a unit. Thus, any integer that can be written as the sum of two squares is not a Gaussian prime.

(f) Verify that 2 is not a Gaussian prime

The integer 2 can be written as $2 = 1^2 + 1^2$. Therefore, it can be factored in the Gaussian integers: $2 = (1+i)(1-i)$.

Let's check the norms of the factors: $N(1+i) = 1^2+1^2=2$ and $N(1-i) = (-1)^2+1^2=2$. Since the norms are not 1, the factors are not units. Thus, 2 has non-trivial factors in $\mathbb{Z}[i]$ and is not a Gaussian prime.

(g) Find another prime of the form $c^2+d^2$ that is not a Gaussian prime

Let's take the prime number 5. We can write $5 = 2^2 + 1^2$. This means $5$ can be factored as $(2+i)(2-i)$. The norms are $N(2+i)=5$ and $N(2-i)=5$. Neither factor is a unit, so 5 is not a Gaussian prime.

(h) Show that $N(\alpha\beta) = N(\alpha)N(\beta)$

Let $\alpha = a+bi$ and $\beta = c+di$. Then $\alpha\beta = (ac-bd) + (ad+bc)i$.

$$N(\alpha\beta) = (ac-bd)^2 + (ad+bc)^2$$

$$= (a^2c^2 - 2abcd + b^2d^2) + (a^2d^2 + 2abcd + b^2c^2)$$

$$= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2$$

Now, let's look at the product of the norms:

$$N(\alpha)N(\beta) = (a^2+b^2)(c^2+d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$$

The expressions are identical. Thus, $N(\alpha\beta) = N(\alpha)N(\beta)$.

(i) Show that 1+4i is a Gaussian prime

First, calculate the norm: $N(1+4i) = 1^2+4^2=17$. If $1+4i$ has a non-trivial factorization $1+4i = (a+bi)(c+di)$, then by the multiplicative property of norms, $N(1+4i) = N(a+bi)N(c+di)$, so $17 = (a^2+b^2)(c^2+d^2)$. Since 17 is a prime number in $\mathbb{Z}$, one of the factors must be 1. For example, $a^2+b^2=1$. This implies that $a+bi$ is a unit. This means any factorization must involve a unit. Therefore, $1+4i$ is a Gaussian prime.

(j) Proof by contradiction: a prime $p$ not of the form $a^2+b^2$ is a Gaussian prime

Suppose a prime $p$ is not of the form $a^2+b^2$ for integers $a, b$. Assume, for the sake of contradiction, that $p$ is not a Gaussian prime. This means $p$ has a non-trivial factorization in $\mathbb{Z}[i]$, so $p=(a+bi)(c+di)$, where neither factor is a unit. Taking norms, we get $N(p) = p^2 = N(a+bi)N(c+di) = (a^2+b^2)(c^2+d^2)$. Since $a+bi$ and $c+di$ are not units, their norms are not 1. Thus, we have a factorization of $p^2$ into two integers greater than 1. This means $a^2+b^2=p$ and $c^2+d^2=p$. But this contradicts our initial assumption that $p$ cannot be written as the sum of two squares. Therefore, the assumption that $p$ is not a Gaussian prime must be false.