Show that $\sqrt{2} \notin \mathbb{Q}(i)$ using infinite descent Standard exercise is to show $\sqrt{2} \notin \mathbb{Q}$ (e.g. Wikipedia).  There are examples on Math.SE such as [1, 2, 3, 4].  
If we adjoin an element to $\mathbb{Q}$ does the same proof by contradiction work?  I'd like to show $\sqrt{2} \notin \mathbb{Q}(i)$ where $x = i$ is a solution to $x^2 + 1 = 0$.  Perhaps I could begin the same way.  We are solving:
$$  p^2 = 2 q^2 $$
where $p, q \in \mathbb{Z}[i]$.  We certainly have $2 = (1+i)(1-i)$ so that $2$ does factor in this new ring, but these two numbers are relatively prime.  We have:
$$ \sqrt{1 \pm i } \notin \mathbb{Q}(i)$$
And therefore, their product does not belong in that field as well.  Does that look correct?  Is has correct ideas, but I don't think the logic is presented correctly.
There are may solutions to this problem, so I've also indicated a certain line of proof I'm trying to follow, using infinite descent.  I'm asking, Does the descent argument we typically use over $\mathbb{Z}$ carries over to $\mathbb{Z}[i]$ ?

Alternatively, we can write any element of $\mathbb{Q}(i)$ as $a + bi$ with $a,b \in \mathbb{Q}$. If we have $(a+bi)^2 = 2$ Then $a^2 - b^2 = 2$ and  $2ab = 0$.  Then necessarily $b = 0$ and $a^2 = 2$.  This only uses descent over $\mathbb{Q}$, and we never use that $2$ factorizes into $(1+i) \times (1-i)$.
In contrast: Proving $\sqrt{2}\in\mathbb{Q_7}$?
 A: Suppose $\sqrt2=a+ib\in\Bbb{Q}[i].$ Then $a^2+b^2=2$ and clearing denominators we can consider the Diophantine equation $p^2+q^2=2r^2.$ Since RHS is even both $p$ and $q$ has to be same parity. Moreover
$$\left(\dfrac{p-q}{2}\right)^2+\left(\dfrac{p+q}{2}\right)^2=\dfrac{p^2+q^2}{2}=r^2.$$
Then using primitive Pythagorean triples we have $p=(m+n)^2-2n^2,$ $q=(m-n)^2-2n^2$ and $r=m^2+n^2$ for some $m,n\in\Bbb{Z}.$ 
If $p=0$ or $q=0,$ then we have a contradiction $\sqrt2\in\Bbb{Q}.$
Otherwise $b\neq 0,$ which is also contradiction as $\sqrt2\in\Bbb{R}.$
A: We want to show that $\mathbb Q(i)$ does not contain a root to $x^2+2=0$.
To do this ione possible route is to this we work in $\mathbb C$ and notice that $\mathbb Q(i)$ must contain $\{a+bi| a,b\in \mathbb Q\}$.
If we can show $\{a+bi| a,b\in \mathbb Q\}$ is a field then clearly we must have $\mathbb Q(i)=\{a+bi| a,b\in \mathbb Q\}$.
Clearly it is closed under addition,multiplication and additive inverses. Recall that $(a+bi)^{-1}=\frac{a-bi}{a^2+b^2}$ which is also of the desired form.
Thus all we have to prove is that $(a+bi)^2\neq -2$ for $a,b\in \mathbb Q$.
Clearly $\sqrt{2}i$ and $-\sqrt{2}i$ are not of this form. (One can also prove that $(a+bi)^2=-2$ has no solution by hand)
