Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*} 
2ab &=  1 \\ 
a^2 + 2b^2 &=  1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*} 
a^2 + 2\left(\frac{1}{2a}\right)^2 &=  1 \\
a^2 + \frac{1}{2a^2} &=  1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation} 
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*} 
\frac{1 \pm i}{2} + 2b^2 &=  1 \\
1 \pm i + 4b^2 &=  2 \\
\pm i + 4b^2 &=  1 \\
4b^2 &=  1 \pm i \\
2b &=  \sqrt{1 \pm i} \\
b &=  \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}
Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?
 A: The truth is that $\sqrt{1+\sqrt2}$ cannot be expressed as $a+b\sqrt2$ where $a,b$ are integers or rationals. If they were reals then the problem becomes trivial.
(Another way to see the first fact above is that the minimal polynomial of $\sqrt{1+\sqrt2}$ is degree-$4$, whereas if it were expressible as $a+b\sqrt2$ with $a,b\in\mathbb Q$ the minimal polynomial would only be quadratic.)
A: Your start of the question Find $\sqrt{1+ \sqrt 2}$ is misleading.
You're supposing that $\alpha = \sqrt{1+ \sqrt 2}$ belongs to the field extension $\mathbb Q(\sqrt 2) / \mathbb Q$... And you proved that it's not the case.
In fact $\alpha$ is an element of degree $2$ over $\mathbb Q(\sqrt 2)$ because $\alpha$ is a root of the polynomial $p \in \mathbb Q(\sqrt 2) [x]$
$$p(x) = x^2 - (1 +\sqrt 2).$$
And $p$ is irreducible over $\mathbb Q(\sqrt 2)$.
A: With a couple of corrections, your answer seems consistent:
$z=\dfrac{1\pm i}2\implies a=\color{red}\pm\sqrt{\dfrac{1\pm i}2}$
$\pm i+4b^2=1\implies 4b^2=1\color{red}\mp i\implies b=\color{red}\pm\dfrac{\sqrt{1\color{red}\mp i}}{2}$
A: Note
\begin{align}
\sqrt{\sqrt{2}+1}&=\sqrt{(\sqrt2-1)(\sqrt2+1)} \cdot\sqrt{\sqrt{2}+1}
= \sqrt{\sqrt{2}-1} (1+\sqrt2)\\
\end{align}
with $a=b = \sqrt{\sqrt{2}-1}$, thus no rational simplification.
