How to proof that 7 is irreducible in the set of integers adjoin root 6 and norm 7 is not prime.

How to proof that $7$ is irreducible $\mathbb{Z}[\sqrt{6}]$, even though $N(7)$ is not prime. 

To solve this, lets look at the rules that govern irreducibles in $\mathbb{Z} [\sqrt{6}]$. 

1) The elements of $\mathbb{Z} [\sqrt{6}]$ are numbers of the form $a+b\sqrt{6}$ with $a,b$ belonging to sets of intergers. 

2) The norm $N(a+b\sqrt{6})$ is $a^2-6b^2$. 

3) The Norm is a multiplicative $N(x,y) =N(x)N(y)$. 

4) The unit i.e invertible elements are precisely the numbers of norm $\pm{1}$. 

5) Norm of $7$ is $49$. 

So if $7=xy$ is a non-trivial factorization, therefore there are no elements of norm $7$.

Indeed, if $a^2-6b^2=\pm{7}$, reduction $\mod{7}$ gives $a^2-6b^2\equiv{0}(\mod7)$ and we can’t have $b\equiv{0}(\mod7)$ and $a^2-6b^2$ would be irreducible by $49$. Observe that we are quietly using the fact that $7$ is prime to conclude $a\equiv{0}$ from $a^2\equiv{0}$. 

Furthermore, because $7$ is prime, the intergers $\mod{7}$ are a field we can divide by the non-zero $b$ to find $(a/b)^2\equiv{6}(\mod{7})$ but $6$ is not anything squared $\mod{6}$ such that the square m$\mod{7}$  are $0,1,2$ and $4$ as you can quickly determine by squaring all residues $\mod{7}$. 

Therefore, the eqn $a^2-6b^2=7$ has no solution in integers, so $7$ is indeed irreducible in $\mathbb{Z}[\sqrt{6}]$.  

Solution by Alon Amit, PhD mathematics, on Quora.