Irreducible but not prime 
In $\mathbb Z[\sqrt{5}]$, $2$ and $1+\sqrt{5}$ are irreducible but not prime.

To  show irreducible
I tried  that there exists $\alpha$ and $\beta$ such that
$$2=\alpha\cdot \beta $$
$$N(2)=N(\alpha)\cdot N(\beta)$$
$$4=N(\alpha)\cdot N(\beta)$$
then there are 3 possibilities (namely $1.4 , 4.1 , 2.2$)
Now,I have to show that $N(\alpha)=2$ is not possible
that is $|a^2-5b^2|=2$ 
$a^2-5b^2=\pm 2$ 
How can I show that this is not possible? and how can i show that $2$ is not prime in $Z[\sqrt{5}]$?
 A: To show that $2$ is not prime, observe that $2$ divides $(\sqrt{5}+1)^2$ but $2$ does not divide $\sqrt{5}+1$. Or else we can use $(\sqrt{5}-1)(\sqrt{5}+1)$. 
Similarly, we can show that $\sqrt{5}+1$ is not prime. For $(\sqrt{5}+1)(\sqrt{5}-1)=4$. So $\sqrt{5}+1$ divides $(20(2)$, but $\sqrt{5}+1$ does not divide $2$. 
To show that $a^2-5b^2$ cannot be equal to $\pm 2$, observe that any square is congruent to $0$, $1$, or $-1$ modulo $5$. Since $a^2-5b^2\equiv a^2\pmod{5}$, we cannot have $a^2-5b^2$ equal to anything congruent to $\pm 2\pmod{5}$. In particular, it cannot be equal to $2$ or $-2$. 
Remark: It turns out that this flaw can be fixed. If we consider the numbers of the form $a+b\sqrt{5}$, where $a$ and $b$ are integers, together with numbers of the form $\frac{a+b\sqrt{5}}{2}$, where $a$ and $b$ are odd integers, we get a structure in which every non-unit irreducible is prime. 
A: You haven't actually defined $N$, but I'll assume from context that
$$N(a + b \sqrt{5}) = a^2 - 5b^2$$
Consider $a^2 - 5b^2 = \pm 2$ modulo $5$; I'll consider the $+2$ case, as the $-2$ case is identical. The equation then reduces to
$$a^2 \equiv 2 \mod{5}$$
But $0^2 \equiv 0$, $1^2 \equiv 4^2 \equiv 1$, and $2^2 \equiv 3^2 \equiv 4$. We conclude that there is no $a$ satisfying the equation, so $N(\cdot) \neq 2$.

In order to show that $2$ is not prime, you must find two numbers $\alpha$ and $\beta$ such that $2 | \alpha\beta$ but $2\nmid \alpha$ and $2 \nmid \beta$. Andre Nicolas' answer suffices here.
