I am not sure what is the right phrase for primes in some ring. The definition I was gave is that in a ring $A$, $p\in A$ is prime if $\forall x,y \in A$ such that $p\mid xy$, $p\mid x$ or $p\mid y$.
I don't know how to start.(I arrived at odd expressions but too long ones) and it is complex for me... I would really appreciate your help.
Suppose that $3=rs$. Then $9=N3=N(rs)=NrNs$. It follows then that say $Nr$ (up to sign) must be $1,3$ or $9$. I claim it cannot be $3$. Note that in such case, we can write $3=a^2+b^2$ for some integers $a,b$. But the squares modulo $4$ are $0,1$; and they add up to $0,1,2$, and $3$ is equivalent to none of $0,1,2$ modulo $4$. More generally, if $p$ is a prime with $p\equiv 3\mod 4$, then $p$ is prime in $\Bbb Z[i]$. Conversely, if $p$ is a prime with $p\equiv 1 \mod 4$, then $p$ is not prime in $\Bbb Z[i]$ (for example, $5=(2+i)(2-i)$ by the square decomposition $5=2^2+1^2$.)
Hint $\,\ 3\mid ab\,\Rightarrow\ 3\mid a\bar a\, b\bar b\,\Rightarrow\, 3\mid a\bar a\,$ or $\,b\bar b\,$ $\Rightarrow$ $\,3\mid a\,$ or $\,b,\ $ by your prior question.
Remark $\ $ Thus primes $\,p\,$ such that $\,p\mid a\bar a \,\Rightarrow\, p\mid a\,$ remain prime in $\,\Bbb Z[i].\,$ Conversely, if $\,p\mid a\bar a\,$ but $\,p\nmid a,\ {\rm\color{#c00}{then}}\,\ p\nmid \bar a ,\,$ so $\,p\,$ is not prime in $\Bbb Z[i].$
${\rm\color{#c00}{Lemma}}\,\ \ b\mid c\iff \bar b\mid \bar c.\ $ Proof: $\,\ ab = c\,\iff \bar a\bar b = \bar c\ $ by multiplicativity of $\,x\mapsto \bar x$
