Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$ Question: Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$
Thoughts: the "standard" method of starting by squaring (twice) to get rid of the square roots, because I don't have a nice way of showing the resulting polynomial is irreducible.  So..
Attempt: It would be great if I could get our $\alpha$ in the form $$(a+b)^2=a^2+2ab+b^2=\sqrt{3+2\sqrt{2}}.$$  So, $$3+2\sqrt{2}=(\sqrt{2})^2+2\sqrt{2}+1=(\sqrt{2}+1)^2.$$So,
$$\alpha=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.$$So,
$$(\alpha-1)^2=2\\
 \alpha^2-2\alpha+1=2\\
\alpha^2-2\alpha-1=0.$$So let $f(x)=x^2-2x-1$.  Since $f(x)$ is irreducible over $\mathbb{Q}$ by the Rational Roots Test (since it has degree $2$), $f(x)$ is monic, and $f(\alpha)=0$, we conclude that $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.
Does this look okay?

OP's proof is good, and there is not much room left to simplify. The following are just alternatives.

*

*Let $\beta = \sqrt{3 - 2\sqrt{2}}$ and note that $\alpha\beta=1$ and $\alpha^2+\beta^2=6$. Given that $0 \lt \beta \lt \alpha$ it follows that $\alpha-\beta=\sqrt{(\alpha-\beta)^2}=\sqrt{\alpha^2+\beta^2-2\alpha\beta}=\sqrt{6-2}=2$.
Then $\alpha(-\beta)=-1$ and $\alpha+(-\beta)=2$, so $\alpha, -\beta$ are the roots of $x^2 - 2 x - 1\,$.


*The "standard" method can also work. Suppose "starting by squaring (twice) to get rid of the square roots", then $x^4 -6x^2+1=0=(x^2-1)^2-4x^2=(x^2-2x-1)(x^2+2x-1)$. Only the first quadratic factor has a root larger than $2$, which must be $\alpha$.
