Gaussian primes are [[Prime Elements of Integral Domain|primes]] in integral domain of [[Gaussian integers|Gaussian Integers]]. >[!NOTE] Theorem 2.3.14 (Classification of Gaussian primes) \[MA257\] > The following is a complete list of Gaussian primes. > - Prime $p \equiv 3 \bmod 4$ times a unit. > - $\alpha \in \mathbb{Z}[i]$ such that $N(\alpha)$ is prime. **Proof**: We've seen that these are Gaussian primes.... Now suppose $\alpha=x+y i$ is a Gaussian prime, where $x, y \in \mathbb{Z}$, and let $p$ be a prime divisor of $N(\alpha)=$ $x^2+y^2$. First suppose $p=2$. Then $x \equiv y \bmod 2$ and $ \alpha=(1+i)\left(\frac{x+y}{2}+\frac{y-x}{2} i\right) $ Consequently, the second factor is a unit, and $N(\alpha)=2$ is prime. Next, suppose instead that $p \equiv 3 \bmod 4$. As $p$ divides $x^2+y^2$, Lemma 2.2 .3 gives $p \mid x$ and $p \mid y$, so $p \mid \alpha$. Thus $\alpha$ is a unit multiple of $p$. Finally, suppose instead that $p \equiv 1 \bmod 4$, and write $p=a^2+b^2$ with $a, b \in \mathbb{N}$. It remains to show that $\alpha$ is divisible by $\pi:=a+b i$ or $\bar{\pi}$ (whereupon $N(\alpha)=p$ is prime). We compute that $ \frac{\alpha}{\pi}=\frac{\bar{\pi} \alpha}{N(\pi)}=\frac{(a-b i)(x+y i)}{a^2+b^2}=\frac{a x+b y}{p}+\frac{a y-b x}{p} i $ and $ \frac{\alpha}{i \bar{\pi}}=\frac{-\pi i \alpha}{N(\pi)}=\frac{-i(a+b i)(x+y i)}{a^2+b^2}=\frac{(b-a i)(x+y i)}{p}=\frac{b x+a y}{p}+\frac{b y-a x}{p} i $ Observe that $ (b y+a x)(b y-a x)=b^2 y^2-a^2 x^2=\left(a^2+b^2\right) y^2-a^2\left(x^2+y^2\right) $ is divisible by $p$, as is $ (a y-b x)(a y+b x)=a^2 y^2-b^2 x^2=\left(a^2+b^2\right) y^2-b^2\left(x^2+y^2\right) $ We now distinguish four cases. Case: $p$ divides $b y+a x$ and $a y-b x$. Then $\pi \mid \alpha$. Case: $p$ divides $b y-a x$ and $a y+b x$. Then $\bar{\pi} \mid \alpha$. Case: $p$ divides $b y+a x$ and $a y+b x$. Then $p$ divides $ y(a x+b y)-x(a y+b x)=b\left(y^2-x^2\right) $ so, as $p$ is too large to divide $b$, it must divide $y^2-x^2$. Since $p$ also divides $y^2+x^2$, we see that $p \mid 2 x^2$ and $p \mid 2 y^2$. Now $p$ divides $x$ and $y$, and so $\pi \mid \alpha$. Case: $p$ divides $b y-a x$ and $a y-b x$. Then $p$ divides $ y(b y-a x)+x(a y-b x)=b\left(y^2-x^2\right) $ so again $\pi \mid \alpha$.