Element of that $\mathbb Q[\sqrt{2}]$ have a square root in $\mathbb Q[\sqrt{2}]$ So I am learning about Quadratic Fields and I have a question:
Consider $\mathbb Q[\sqrt{2}]$. So does every element of that $\mathbb Q[\sqrt{2}]$ have a square root in $\mathbb Q[\sqrt{2}]$. I think that this essentially means that . I don't know if this is right though. Is this true or false. If true why is it true and if falso can someone show me a counterexample. 
 A: Let $\alpha = a+b\sqrt{2} \in \mathbb{Q}[\sqrt{2}]$ be such that $\alpha^2=\sqrt{2}$. Then we get
\begin{align*}
(a+b\sqrt{2})^2 & = \sqrt{2}\\
a^2+2b^2+2ab\sqrt{2} & = \sqrt{2}.
\end{align*}
This yields
\begin{align*}
a^2+2b^2 & = 0\\
2ab & = 1.
\end{align*}
But this has no solutions for $a,b \in \mathbb{Q}$. Thus there is no square root of $\sqrt{2}$ inside $\mathbb{Q}[\sqrt{2}]$.
A: If we had $\sqrt{\sqrt2}\in\mathbb Q[\sqrt2]$, it would follow that there are integers $l,m,n$ such that $l\sqrt{\sqrt2}=m+n\sqrt2$. By squaring, we could conclude that $$\sqrt2=\frac{m^2+2n^2}{l^2-2mn}\in \Bbb Q.$$
A: I think the question is asking that if you have $\alpha^2 = c + d\sqrt{2} $, for ANY $c$ and $d$, then $\alpha$ can be represented as $a + b\sqrt{2}$.
Here's what I came up with:
Let $\alpha = a + b\sqrt{2} \in Q[\sqrt{2}]$.
$$\alpha^2 = (a + b\sqrt{2})^2 = a^2+2ab\sqrt{2}+2b^2=(a^2+2b^2)+(2ab)\sqrt{2} $$
Let $c = a^2+2b^2$ and $d=2ab$, then $\alpha^2= c + d\sqrt{2}$.
From this we can see that in order for $\sqrt{a_1+b_1\sqrt{2}} = a_2+b_2\sqrt{2}$, then $a_1=a_2^2+2b_2^2$ and $b_1=2a_2b_2$.
One example that works is $a_2=\frac{2}{3}$ and $b_2=\frac{4}{7}$, then $a_1 = \frac{484}{441}$ and $b_1 = \frac{16}{21}$. Indeed, if we plug in $\sqrt{\frac{484}{441}+\frac{16}{21}\sqrt{2}}$, we get $\frac{2}{3}+\frac{4}{7}\sqrt{2}$.
However, a counter example would be one that doesn't follow the formulas. For example, if we have $\sqrt{\frac{483}{441}+\frac{16}{21}\sqrt{2}}$ (the numerator of the "a" term was changed), it has no rational square roots of the form $a+b\sqrt{2}$
$\square$
