I have an idea that:
Let $c\in \mathbb Z$, $c$ is reducible in $\mathbb Z[i]$ iff $\,$$|\,c\,|$ is a sum of square of two integers (include 0).
I only need a quite answer to my hypothesis. Whether it is true or not. If it is false, then please show me a counter example.
Reduciblility is in a ring sense, i.e. in a commutative ring $R$, if $a$ is reducible in $R$, then there exists an equation $a=mn$, s.t. $m\notin R^*$ and $n\notin R^*$. Here $R^*$ means the set of unit of $R$.
Also, I notice that let $c\in \mathbb Z$ is quite limited. Is there a pattern in $\mathbb Z[i]$ for those reducible elements? Or a theorem relate to the reducible elements in Guassian Integers.
This is false.
Take $c=6$. Then $c=2\cdot 3$ and so $c$ is reducible in $\mathbb Z[i]$ because neither $2$ nor $3$ is a unit in $\mathbb Z[i]$. However $6$ is not a sum of two squares in $\mathbb Z$.
Take $c=1$. Then $c$ is a sum of two squares in $\mathbb Z$ but is a unit in $\mathbb Z[i]$, so not reducible.
If you eliminate $c=\pm 1$, then it is true that $c$ is reducible in $\mathbb Z[i]$ if $|c|$ is a sum of square of two integers in $\mathbb Z$. Indeed, $|c|=a^2+b^2$ implies $\pm c=(a+bi)(a-bi)$.
The irreducible elements in $\mathbb Z[i]$ are exactly the primes $q \in \mathbb Z$ of the form $4n+3$ or the numbers $a+bi$ such that $a^2+b^2$ is a prime (which will be $2$ or a prime of the form $4n+1$).