## Exhibit some irreducible Gaussian integers

Find all the irreducible elements $a + bi \in \mathbb{Z}[i]$, where $|a|, |b| \leq 5$.

Note that if $\alpha$ is irreducible, then all the associates of $\alpha$ are also irreducible, as well as their conjugates. Thus from each irreducible Gaussian integer we can easily construct seven others. We claim that if $\alpha$ is a Gaussian integer, then one of these eight is of the form $a+bi$ where $0 \leq a \leq b$. To see this, let $\alpha = a+bi$ and note the following.

• If $0 \leq a \leq b$, then $a+bi$ has this form.
• If $0 \leq b \leq a$, then $b + ai$ has this form.
• If $a \leq 0 \leq b$ and $|a| \leq |b|$, then $(-a) + bi$ has this form.
• If $a \leq 0 \leq b$ and $|a| \geq |b|$, then $b + (-a)i$ has this form.
• If $b \leq 0 \leq a$ and $|a| \leq |b|$, then $a + (-b)$ has this form.
• If $b \leq 0 \leq a$ and $|a| \geq |b|$, then $(-b) + ai$ has this form.
• If $a \leq b \leq 0$, then $(-b) + (-a)i$ has this form.
• If $b \leq a \leq 0$, then $(-a) + (-b)i$ has this form.

So every Gaussian integer is conjugate-associate to one of the form $a+bi$ where $0 \leq a \leq b$. Suppose $a = b \neq 1$; then $a+bi = a(1+i)$ is reducible. Thus every irreducible Gaussian integer is conjugate-associate to either $1+i$ or to $a+bi$ where $0 \leq a < b$.

To find the irreducible Gaussian integers $a+bi$ satisfying $|a|,|b| \leq 5$, it suffices to consider the following 16 cases.

1. $i$: This is a unit, and so not irreducible.
2. $2i$: Note that $2i = (1+i)^2$ has a nontrivial factorization, and so is not irreducible.
3. $3i$: Note that $N(3i) = 9$. Suppose $N(x+yi) = 3$, so that $x^2+y^2 = 3$. This equation has no solutions in the integers. Thus in any factorization of $3i$, some factor must have norm 1, and thus must be a unit. Hence $3i$ is irreducible.
4. $4i$: Note that $4i = (2)(2i)$ has a nontrivial factorization, and so is not irreducible.
5. $5i$: Note that $5i = (1+2i)(2+i)$ has a nontrivial factorization, and so is not irreducible.
6. $1+i$: Note that $N(1+i) = 2$ is prime, so that this element is irreducible.
7. $1+2i$: Note that $N(1+2i) = 5$ is prime, so that this element is irreducible.
8. $1+3i$: Note that $1+3i = (1+i)(2+i)$ has a nontrivial factorization, and so is not irreducible.
9. $1+4i$: Note that $N(1+4i) = 17$ is prime, so that this element is irreducible.
10. $1+5i$: Note that $1+5i = (i-1)(2-3i)$ has a nontrivial factorization, and so is not irreducible.
11. $2+3i$: Note that $N(2+3i) = 13$ is prime, so that this element is irreducible.
12. $2+4i$: Note that $2+4i = 2(1+2i)$ has a nontrivial factorization, and so is not irreducible.
13. $2+5i$: Note that $N(2+5i) = 29$ is prime, so that this element is irreducible.
14. $3+4i$: Note that $3+4i = (2i-1)(1-2i)$ has a nontrivial factorization, and so is not irreducible.
15. $3+5i$: Note that $3+5i = (4i-1)(1-i)$ has a nontrivial factorization, and so is not irreducible.
16. $4+5i$: Note that $N(4+5i) = 41$ is prime, so that this element is irreducible.