# Degree of the extension $\mathbb{Q}(\sqrt{7}+\sqrt{49})$

The original question is find the degree of the irreducible polynomial of $$3+\sqrt{7}+\sqrt{49}$$, but it's equivalent to find $$[\mathbb{Q}(3+\sqrt{7}+\sqrt{49}):\mathbb{Q}]$$.

$$\mathbb{Q}(3+\sqrt{7}+\sqrt{49})=\mathbb{Q}(\sqrt{7}+\sqrt{49})=\mathbb{Q}(\sqrt{7}+(\sqrt{7})^{2})$$. But i don't know how to continue. If the question was $$\mathbb{Q}(\sqrt{7}+(\sqrt{7})^{2})$$: $$\alpha=\sqrt{7}+(\sqrt{7})^{2}$$ $$\alpha^{3}=7+3*7\sqrt{7}+3*7(\sqrt{7})^{2}+49$$ $$\alpha^{3}=21\alpha+56$$ $$X^{3}-21X-56$$ Irreducible, because Eisenstein's criterion with $$p=7$$. But this method doesn't work with $$n=5$$, or at least it isn't clear to find the relation. I think there should be another method to solve this more easily.

Since $$\mathbb Q\subseteq \mathbb{Q}(\sqrt{7}+\sqrt{49}) \subseteq \mathbb{Q}(\sqrt{7})$$ and $$[\mathbb{Q}(\sqrt{7}):\mathbb Q]=5$$, we must $$[\mathbb{Q}(\sqrt{7}+\sqrt{49}):\mathbb Q] =1$$ or $$5$$. But $$\sqrt{7}+\sqrt{49}$$ is not rational, for otherwise $$\sqrt7$$ will be the root of some polynomial of the form $$x^2+x-q$$, where $$q\in\mathbb Q$$, contradicting the fact that $$x^5-7$$ is irreducible.
Hint. Let us call your element $$\alpha$$. It is in the field $$\mathbf{Q}(\sqrt{7})$$, which is of degree $$5$$ over $$\mathbf{Q}$$. As $$5$$ is prime, it suffices by the tower law to prove that $$\alpha\not\in \mathbf{Q}$$ to conclude that $$\alpha$$ also has degree $$5$$ over $$\mathbf{Q}$$.