<s>[INST] Constructing a degree 4 rational polynomial satisfying $f(\sqrt{2}+\sqrt{3}) = 0$ Goal: Find $f \in \mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3}) = 0$.
A direct approach is to look at the following
$$
\begin{align}
(\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\
(\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\
\end{align}
$$
Putting those together gives
$$
-1 + 10(\sqrt{2}+\sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^4 = 0,
$$
so $f(x) = -1 + 10x^2 - x^4$ satisfies $f(\sqrt{2}+\sqrt{3}) = 0$.
Is there a more mechanical approach? Perhaps not entirely mechanical, but something more abstract.
 [/INST] A 'mechanical' approach follows. Let $x=\sqrt{2}+\sqrt{3}$. Then $x^2=5+2\sqrt{6}$ which means $x^2-5=2\sqrt{6}$. Now $$(x^2-5)^2=24\Longrightarrow(x^2-5)^2-24=0.$$ By construction, one of the roots of $f(x)=(x^2-5)^2-24$ is $\sqrt{2}+\sqrt{3}$.
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