Question: Let $\mathbb{Z}[i]$ denote the Gaussian integers.
(a) Compute the norm $N(3+i)$ of $3+i$ in $\mathbb{Z}[i]$
(b) Factor both $3+i$ and its norm into primes in $\mathbb{Z}[i]$
(c) Compute $\gcd_{\mathbb{Z}[i]}\left(3+i,2\right)$.
For (a), we have $N(3+i)=3^2+1^2=10$.
For (b), I would suspect the following to be useful
Proposition: The prime elements in $\mathbb{Z}[i]$ are as follows:
(a) $1+i$
(b) the primes $p\in\mathbb{Z}$ with $p \equiv 3 \bmod{4}$, and
(c) $a \pm bi$ where $p=a^2+b^2=(a+bi)(a-bi)$ for the primes $p\in\mathbb{Z}$ with $p \equiv 1 \bmod{4}$.
However, I don't see how the above helps with factoring, for example, the norm value $10$. To me, the only logical thing to write is $10 = 2\cdot5$. Note that $5=2^2+1^2=(2+i)(2-i)$ and $5\in\mathbb{Z}$ is a prime with $5\equiv 1 \bmod{4}$. Therefore, by the above proposition, $(2+i)$ and $(2-i)$ are prime, and we obtain $10 = 2 \cdot (2+i) \cdot (2-i)$. But what to do about the $2$? I could write $2=(1+i)(1-i)$, but I don't believe $1-i$ is prime.
For $3+i$, I found that $3+i=(1+i)(2-i)$. Now $1+i$ is prime by (a). Since $5=2^2+1^2=(2+i)(2-i)$ and $5\in\mathbb{Z}$ is prime with $5 \equiv 1 \bmod{4}$, we have that both $(2+i)$ and $(2-i)$ are prime, in particular $(2-i)$. Therefore $(1+i)(2-i)$ factors $3+i$ into its primes in $\mathbb{Z}[i]$.
For (c), I found that $\gcd_{\mathbb{Z}[i]}\left(3+i,2\right)=-1-i$. Is this correct?