# Equivalence of Definitions of Gaussian Prime

## Theorem

The following definitions of the concept of **Gaussian Prime** are equivalent:

### Definition $1$

Let $x \in \Z \sqbrk i$ be a Gaussian integer.

$x$ is a **Gaussian prime** if and only if:

- it cannot be expressed as the product of two Gaussian integers, neither of which is a unit of $\Z \sqbrk i$ (that is, $\pm 1$ or $\pm i$)
- it is not itself a unit of $\Z \sqbrk i$.

### Definition $2$

A **Gaussian prime** is a Gaussian integer which has exactly $8$ divisors which are themselves Gaussian integers.

## Proof

Let $x = a + b i$ be a Gaussian integer.

We have:

\(\ds x\) | \(=\) | \(\ds 1 \times \paren {a + b i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds -1 \times \paren {-a - b i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds i \times \paren {b - a i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds -i \times \paren {-b + a i}\) |

Thus it is seen that every Gaussian integer has $8$ divisors which are themselves Gaussian integers.

### Definition $(1)$ implies Definition $(2)$

Let $x$ be a Gaussian prime by definition $1$.

Then by definition:

- $x$ cannot be expressed as the product of two Gaussian integers, neither of which is a unit (that is, $\pm 1$ or $\pm i$).

That is, the only divisors of $x$ are those $8$ which have been identified above.

Thus $p$ is a Gaussian prime by definition $2$.

$\Box$

### Definition $(2)$ implies Definition $(1)$

Let $p$ be a Gaussian prime by definition $2$.

Then by definition $p$ has exactly $8$ divisors which are Gaussian integers.

Those are the ones given above.

As those are the only ones, $p$ cannot be the product of two Gaussian integers such that both of them are not a unit.

Thus $p$ is a Gaussian prime by definition $1$.

$\blacksquare$