If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a root in $\mathbb{Q}(\sqrt{2})$, then the root is actually in $\mathbb{Z}[\sqrt{2}]$ I'm working my way though a classical geometry book by Hartshorne right now, but this problem popped up in a section I'm reading. It is Problem 13.10 from Hartshorne's Geometry: Euclid and Beyond if you're curious.
Anyway, the problem states:

If $a,b\in\mathbb{Z}$, and if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$, then the square root is actually in $\mathbb{Z}[\sqrt{2}]$. 

I'm not super familiar with algebra, so I'm having trouble interpreting the question, but I would like to know how to solve it.
I looked up $\mathbb{Q}(\sqrt{2})$ on wikipedia, and it seems that it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Q}\}$. I couldn't find  $\mathbb{Z}[\sqrt{2}]$, but I assume it is the set $\{a+b\sqrt{2}\ |\ a,b\in\mathbb{Z}\}$. 
So if $a+b\sqrt{2}$ has a square root in $\mathbb{Q}(\sqrt{2})$ means there exists some $c+d\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ such that 
$$
(c+d\sqrt{2})^2=c^2+2d^2+2cd\sqrt{2}=a+b\sqrt{2}.
$$
This implies (I think?) that $c^2+2d^2=a$ and $2cd=b$. If this is the correct path, is there then someway to conclude that $c$ and $d$ are in fact integers? Thanks.
By the way, is this exercise easily related to some aspect of classical geometry? It seems kind of out of the blue to me.
 A: First of all $\mathbb{Q}(\sqrt{2})$ is the smallest field containing $\mathbb{Q}$ and $\sqrt{2}$, where $\mathbb{Z}[\sqrt{2}]$ is the smallest ring containing $\mathbb{Z}$ and $\sqrt{2}$. (That is square brackets mean ring, parentheses mean field).
From what you have, I am confident you can conclude that $c$ and $d$ are integers.  Here's what i would do to find it directly:
Let $c=\frac{p}{q}, d=\frac{r}{s}$ with $gcd(p,q)=1$ and $gcd(r,s)=1$.  Since $cd=\frac{pr}{qs}$ is an integer, this implies that $qs$ divides $pr$ and because of the conditions $gcd(p,q)=1$ and $gcd(r,s)=1$, $qm=r$ and $sn=p$, with $n$ and $m$ integers.
And we know $a=\frac{s^2n^2}{q^2}+2\frac{q^2m^2}{s^2}=\frac{s^4n^2+2q^4m^2}{q^2s^2}$ is an integer.  Hence $q^2s^2$ divides $s^4n^2+2q^4m^2$.  Since $q^2$ divides the second term and the whole thing, it must divide the first term, and the same with $s^2$.  But since $q^2$ and $s^2n^2$ are relatively prime, this implies $q^2$ is 1, and the same follows for $s$.
I'm fairly certain there is a cleaner way to do this using the Gauss Lemma, but this is something you can work out directly, which is also nice.
A: First $\rm\ 2\: c^2\: $ times $\rm\ c^2+2\ d^2 =\: a\ $ yields $\rm\ 2\: c^4 + b^2 =\: 2\: a\  c^2\ $ hence $\rm\ 2\: c\in \mathbb Z\ $ by the Rational Root Test. 
Next $\:4\ $ times $\rm\ 2\ d^2 = \:a-c^2\: \ \to\ \ 8\ d^2 = \:4\ a\ - (2\:c)^2 \in \mathbb Z\ \:$ thus $\rm\: 2\: d\in \mathbb Z\ $
Finally $\rm\:  4\ a - 2\ (2\: d)^2 \:=\: (2\:c)^2\: \Rightarrow\ 2\:|\:(2\ c)^2\Rightarrow 2\: |\: 2\: c\ \Rightarrow\ c\in \mathbb Z\ \Rightarrow\ d\in \mathbb Z\quad\quad$ QED
