Show that $3$ is not a prime in $\mathbb Q[\sqrt{7}]$ Question:

Show that $3$ is not a prime in $\mathbb Q [\sqrt 7] $.

To show this, should I start by assuming that $3 = ab$ where $a$ and $b$ are integers in $\mathbb Q[\sqrt{7}]$ and then try to show they are not units? 
What should I do to show that? Or is there another better way to do this problem?
 A: If 3 is prime in this ring, the equations $x^2 + bx \pm 3 = 0$ would have no solutions for $x \in \mathbb Z[\sqrt 7]$, where $b \in \mathbb Z$.
To use the quadratic formula, we have $a = 1$ and $c = \pm 3$.
$$x = \frac{-b \pm \sqrt{b^2 \pm 12}}{2}$$
We could hope that $b$ is small, and in this case it is, but it's not very smart to try several potential values by brute force.
However... since we know we want the solution to have $\sqrt 7$, we just need to solve $b^2 \pm 12 = 28 (= 4 \times 7)$ in integers. The only solutions are $b = \pm 4$ for $b^2 + 12 = 28$ (since 28 + 12 = 40, which is not a square).
Then $$x = \frac{\pm 4 \pm \sqrt{28}}{2} = \pm 2 \pm \sqrt 7$$ (since $\sqrt{28} = 2 \sqrt 7$). None of these numbers are units, because if they were, they'd be solutions to $x^2 \pm 4x \pm 1 = 0$, which clearly, by the foregoing, they're not.
Lastly, just to make sure we haven't made any silly arithmetic mistakes along the way, we check that $$(2 - \sqrt 7)(2 + \sqrt 7) = -3.$$
A: $$3 = (\sqrt 7 - 2)(\sqrt 7 + 2)$$
And  $2^2 - 1^2 \dot \, 7 \neq \pm4 $ or $2^2 - 1^2 \dot\, 7 \neq 1$.
Edit: A fundamental unit in $\mathbb Q[\sqrt 7]$ is $8 + 3\sqrt 7$. Therefore the units are of the form $(8 + 3 \sqrt 7)^n , n\in \mathbb Z$. 
A: Recall that in a commutative ring $R$, an ideal $I$ is prime iff $R/I$ is a domain.  Also note that $\mathbb{Z}[\sqrt{7}] = \frac{\mathbb{Z}[x]}{(x^2-7)}$.  Then
\begin{align*}
\frac{\mathbb{Z}[\sqrt{7}]}{(3)} &\cong \frac{\mathbb{Z}[x]/(x^2-7)}{(3,x^2-7)/(x^2-7)} \cong \frac{\mathbb{Z}[x]}{(3,x^2-7)} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x^2-7)} = \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x^2-1)}\\
&=\frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)(x+1)} \cong \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x-1)} \times \frac{(\mathbb{Z}/3\mathbb{Z})[x]}{(x+1)} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}
\end{align*}
where the second isomorphism holds by the Third Isomorphism Theorem, and the second-to-last by the Chinese Remainder Theorem.  Since $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ is not a domain, then $(3)$ is not prime in $\mathbb{Z}[\sqrt{7}]$.
Moreover, we can recover a factorization of $3$ from this isomorphism.  Since
$$
x^2 - 7 \equiv x^2 - 1 = (x+1)(x-1) \equiv (x-2)(x+2) \pmod{3}
$$
replacing $x$ by $\sqrt{7}$ yields the factorization $3 = (\sqrt{7}-2)(\sqrt{7}+2)$.
A: I believe 3 is small enough that you can just use the continued fraction digits of $\sqrt{7} = [2;\overline{1,1,1,4}]$.
The approximations are $2, 3, \frac{5}{2}, \frac{8}{3},\frac{37}{14},\dots$ Indeed $37^2 - 7\cdot 14^2 = -3$ and even $2^2 - 7*1^2 = -3$.  
These lead to factorizations $3 = (\sqrt{7}-2)(\sqrt{7}+2) =(14\sqrt{7}-37)(14\sqrt{7}+37) $
