Prove 3 is prime in the ring $\Bbb{Z[i]}$. 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.

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 $\,x\mapsto \bar x$
