Factoring 2 in $\mathbb{Q}(\sqrt{-7})$ $7$ is a Heegner number, so all numbers in $\mathbb{Q}(\sqrt{-7})$ have a unique factorization.  
I'm told that:   


*

*$2$ is not prime in $\mathbb{Q}(\sqrt{-7})$,

*$3$ is not prime in $\mathbb{Q}(\sqrt{-11})$,  

*$5$ is not prime in $\mathbb{Q}(\sqrt{-19})$,  

*$11$ is not prime in $\mathbb{Q}(\sqrt{-43})$,  

*$17$ is not prime in $\mathbb{Q}(\sqrt{-67})$,  

*$41$ is not prime in $\mathbb{Q}(\sqrt{-163})$.   


For the first of these, I've tried finding prime numbers $(a+b (1+\sqrt{-7})/2)$ and $(c+d (1+\sqrt{-7})/2)$, with product $2$, but I haven't had luck so far.
 A: $$2=\frac{1+\sqrt{-7}}{2}\frac{1-\sqrt{-7}}{2}$$
$$3=\frac{1+\sqrt{-11}}{2}\frac{1-\sqrt{-11}}{2}$$
$$5=\frac{1+\sqrt{-19}}{2}\frac{1-\sqrt{-19}}{2}$$
$$11=\frac{1+\sqrt{-43}}{2}\frac{1-\sqrt{-43}}{2}$$
See the pattern ?
A: Since 2 is not pure imaginary you must, for non zero $b,d$, have $a=c$ and $d=-b$. This reduces to solving $(\frac{a+b\sqrt{-7}}{2})(\frac{a-b\sqrt{-7}}{2})=2$, so you need to find $a,b$ such that $a^2+7b^2=8$.
A: Your mileage may vary...
I find these problems much easier if instead of trying to solve $(a - b \theta)(a + b \theta)$ (where $\theta = \frac{1 + \sqrt d}{2}) = p$, I try to solve $$\left( \frac{a - b \sqrt d}{2} \right) \left( \frac{a + b \sqrt d}{2} \right) = p.$$ Then $$\left( \frac{a - b \sqrt d}{2} \right) \left( \frac{a + b \sqrt d}{2} \right) = \frac{a^2 + (-d)b^2}{4}$$ and $a^2 + (-d)b^2 = 4p$.
So then, for $d = -7$, $p = 2$, I solve $a^2 + 7b^2 = 8$. The answer then becomes obvious: $a = 1$, $b = 1$ also.
With the  thetas, the whole thing is kind of confusing.
$$2 = \frac{1 + 7}{4} = \left( \frac{1 + \sqrt{-7}}{2} \right) \left( \frac{1 - \sqrt{-7}}{2} \right) = (1 - \theta) \theta,$$
(adjust $\theta$ as you go on to each of these)
$$3 = \frac{1 + 11}{4} = \left( \frac{1 + \sqrt{-11}}{2} \right) \left( \frac{1 - \sqrt{-11}}{2} \right) = (1 - \theta) \theta,$$
$$5 = \frac{1 + 19}{4} = \left( \frac{1 + \sqrt{-19}}{2} \right) \left( \frac{1 - \sqrt{-19}}{2} \right) = (1 - \theta) \theta,$$
$$11 = \frac{1 + 43}{4} = \left( \frac{1 + \sqrt{-43}}{2} \right) \left( \frac{1 - \sqrt{-43}}{2} \right) = (1 - \theta) \theta,$$
which is to say, for your $(a + b \theta)(c + d \theta)$, we have $a = 1$, $b = -1$, $c = 0$, $d = 1$. The thetas obscure your approach to $4p$. But, like I said, your mileage might vary.
A: As has been noted in comments, $2$ is a unit in $\mathbb Q[\sqrt{-7}]$, since $\frac12\cdot 2 = 1 $ and $\frac12\in\mathbb Q\subseteq\mathbb Q[\text{anything}]$. Thus, its factorization is trivial.
$2$ is irreducible in $\mathbb Z[\sqrt{-7}]$ (to see this, note that every nonzero element has complex magnitude $\ge 1$, and the only elements with magnitude smaller than $2$ are $\pm 1$, which are units).
As Rene Schipperus points out, $2$ does factor in the ring of algebraic integers in $\mathbb Q[\sqrt{-7}]$, which consists of the numbers
$$ \Bigl\{ (a+m)+(b+m)\sqrt{-7} ~\Bigm|~ a,b\in \mathbb Z, m\in\{0,\tfrac12\} \Bigr\} $$
