Finding a polynomial with integer coefficients that has $\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ as a root

Question

I'm looking for an (ideally) elementary way to find a polynomial with integer coefficients that has $$\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$$ as a root.

What I have so far is that we can write $\sqrt 3 - \sqrt 2$ as $1/(\sqrt 3 + \sqrt 2)$ and then we can quite easily check that both $\sqrt 3 + \sqrt 2$ and $\sqrt 3 - \sqrt 2$ solve $$x^4-10x^2+1=0.$$

So then we have that $\sqrt[100]{\sqrt{3} + \sqrt{2}}$ and $\sqrt[100]{\sqrt{3} - \sqrt{2}}$ both solve $$x^{400}-10x^{200}+1=0$$

but I am not sure how to find an integer-coefficient polynomial with the sum of those two values as a root. It sort of looks like Vieta formulas could somehow be relevant, but I may be missing something obvious here.

Answer 1

Hint

Denote

$$\begin{cases} a &= \sqrt[100]{\sqrt 3 + \sqrt 2}\\ b &= \sqrt[100]{\sqrt 3 - \sqrt 2} \end{cases}$$

and notice that $ab=1$. We have $a^{100} + \frac{1}{a^{100}} = 2\sqrt 3$.

Now you can write

$$a^{100} + \frac{1}{a^{100}} = \left(a + \frac{1}{a}\right)^{100} - (p(a)+p(1/a))$$ where $p(x)$ is a polynomial of degree at most equal to $99$ with integer coefficients. You can repeat this process by induction to finally write $a^{100} + \frac{1}{a^{100}} = q(a+1/a)$ where $q$ is a polynomial of degree $100$ with integer coefficients.

At the end you get

$$q(a+b) =2 \sqrt 3$$ which implies that $a+b$ can't be a rational number as $\sqrt 3$ isn't. And that $a+b$ is a root of the polynomial with integer coefficients $$q^2(x) - 12=0$$ as desired.