Is $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$ true? Motivated by the positive answer to the following question:  
Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?
I'm curious about whether ${\Bbb Z}[\sqrt{2}+\sqrt{3}]=\Bbb{Z}[\sqrt{2},\sqrt{3}]$ is also true, where ${\Bbb Z}[\alpha]$ denotes the smallest subring of ${\Bbb C}$ which contains $\alpha\in{\Bbb C}$. It suffices to know whether $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$ is true or not. With some manipulation one can get
$$
2\sqrt{2}\in {\Bbb Z}[\sqrt{2}+\sqrt{3}]. 
$$
It seems no hope to get $f(x)\in\Bbb{Z}[x]$ such that
$$
f(\sqrt{2}+\sqrt{3})=\sqrt{2},
$$
which is equivalent to $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$. But I don't have a proof. How should I go on?
 A: Without being a 'pretty' solution, it's not too difficult to show that even powers of $\sqrt{2}+\sqrt{3}$ are of the form $a+b\sqrt{6}$ with $a,b\in {\Bbb Z}$, and odd powers are of the form $a\sqrt{2}+b\sqrt{3}$ with $a,b\in{\Bbb Z}$ and $a,b$ both odd.
No matter how many of the latter you add up, you'll never have the parity of the coefficients differing.
A: Hint $\ $ Note $\,\ \sqrt{b}\,\not\in \Bbb Z[\alpha]\ $ for  $\ \alpha\, =\, \sqrt b +\! \sqrt{a}\ $  of degree $\,4\,$ over $\Bbb Q,\,$ else
$$\!\!\!\! \begin{eqnarray}
\alpha\,(2\sqrt b-\!\alpha)&=&\phantom{._{I^{I^I}}}\!\!\!\!\!\!\!\!\!\! (\sqrt b+\!\sqrt a)(\sqrt b-\!\sqrt a)\, =\, \color{#0a0}{b\!-\!a}\\
\Rightarrow\ \  \alpha\sqrt b\,  =\, \dfrac{\alpha^2}{\color{#c00}2}\!&+&\!\dfrac{\color{#0a0}{b\!-\!a}}2\,\in\,\color{#c00}{\Bbb Z}[\alpha]\, =\,\color{}{\Bbb Z}\!+\!\alpha\Bbb Z\!+\!\color{}{\alpha^2{\color{#c00}{\Bbb Z}}}\!+\!\alpha^3\Bbb Z \,\ \Rightarrow\ \dfrac{1}{\color{#c00}2} \in \color{#c00}{\Bbb Z}\end{eqnarray}$$
A: Letting $\alpha = \sqrt{2} + \sqrt{3}$, we have $\sqrt{2} = \frac{1}{2}\alpha^3 - \frac{9}{2}\alpha \in \mathbf{Q}(\alpha)$.
The minimal polynomial of $\alpha$ over $\mathbf{Q}$ is $f(x) = x^4 - 10x^2 + 1$. Thus any element $z$ of $\mathbf{Q}(\alpha)$ has a unique representation of the form $z = a\alpha^3 + b\alpha^2 + c\alpha + d$ with $a, b, c, d \in \mathbf{Q}$. Moreover, because $f(x)$ has integer coefficients, we have $z \in \mathbf{Z}[\alpha]$ if and only if $a, b, c, d \in \mathbf{Z}$.
It follows that $\sqrt{2} \not\in \mathbf{Z}[\alpha]$.
