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 $1$.
$\blacksquare$