Factorise $13$ into a product of irreducibles in $\Bbb Z[i]$ I need to factorise $13$ into a product of irreducibles in $\Bbb Z[i]$ but I'm having trouble factorising it at all.
So far I have
$13=(a+bi)(c+di)$
$13=ac + cbi + adi - bd$
So
$bc = -ad$
$ac-bd=13$
I don't know how to go any further for working out the factors for it. Any direction would be appreciated.
 A: $13=2^2+3^2=(2-3i)(2+3i)$
General result: If $p$ is a prime and 
$p \equiv 1$ mod $4$ then $p=(a-bi)(a+bi)$ for some $a,b \in \mathbb{Z}$ and $(a-bi),(a+bi)$ are irreducible in $\mathbb{Z}[i]$ .
You can see the proof of this result in the page 290, ABSTRACT ALGEBRA , authors: David S.Dummit and Richard M.Foot.
A: Once you know the definition of a prime number in $\mathbb{Z}[i]$ and a fact about about Gaussian integers this problem becomes much easier:
Definition: A number $a+bi \in \mathbb{Z}[i]$ is a Gaussian prime iff 
1) $a^2+b^2$is a prime in $\mathbb{N}$.
2) $a=0$ and $b$ is prime with remainder of 3 when divided by 4.$(b \equiv 3 \; mod 4)$.
3) $b=0$ and $a$ is prime with remainder of 3 when divided by 4.$(a \equiv 3 \; mod 4)$.
The norm in the complex plane is multiplicative. So, for two complex integers $z_0,z_1$, we have $N(z_0z_1)=N(z_0)N(z_1)$. 
Gaussian primes are irreducible in $\mathbb{Z[i]}$, so start there. The number $13$ is a prime, but $13 \equiv 1 \; mod 4$.This means it is not a Gaussian prime, so it factors.
Now we use the multiplicative property of the norm. $N(13)=13^2+0^2=13*13=N(z_0)N(z_1)$, where $z_0,z_1$ are factors of 13 in $\mathbb{Z[i]}$. Let $z_0=a+bi$, then $N(a+bi)=a^2+b^2=13$.
At this point it should be clear what $a,b$ are. Since you want a positive integer and both Gaussian integers have the same norm, once you find one of the factors the other must be the conjugate. $z_1=\overline{z_0}$
A: Hint: Note that $13=(3+2i)(3-2i)$. To show these are irreducible, use the fact that the norm of a product is the product of the norms. Please note that here norm is the number-theoretic version: the norm of $a+bi$ is $a^2+b^2$. 
