Is $\sqrt{2}\in{\Bbb Z}[\sqrt{2}+\sqrt{3}]$ true?

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 $\sqrt{2} \not\in \mathbf{Z}[\alpha]$.

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.

Hint $\ $ Note $\,\ \sqrt{b}\,\not\in \Bbb Z[\alpha]\ $ for $\ \alpha\, =\, \sqrt b +\! \sqrt{a}\ $ of degree $\,4\,$ over $\Bbb Q,\,$ else $$\!\!\!\! \begin{eqnarray} \alpha\,(2\sqrt b-\!\alpha)&=&\phantom{._{I^{I^I}}}\!\!\!\!\!\!\!\!\!\! (\sqrt b+\!\sqrt a)(\sqrt b-\!\sqrt a)\, =\, \color{#0a0}{b\!-\!a}\\ \Rightarrow\ \ \alpha\sqrt b\, =\, \dfrac{\alpha^2}{\color{#c00}2}\!&+&\!\dfrac{\color{#0a0}{b\!-\!a}}2\,\in\,\color{#c00}{\Bbb Z}[\alpha]\, =\,\color{}{\Bbb Z}\!+\!\alpha\Bbb Z\!+\!\color{}{\alpha^2{\color{#c00}{\Bbb Z}}}\!+\!\alpha^3\Bbb Z \,\ \Rightarrow\ \dfrac{1}{\color{#c00}2} \in \color{#c00}{\Bbb Z}\end{eqnarray}$$