Messing around, I noticed that


$$\sqrt{2}=e^{i\pi /4}+e^{-i\pi /4}$$


$$\sqrt{3}=e^{i\pi /6}+e^{-i\pi /6}$$


and (even more surprisingly)


$$\sqrt{5}=e^{2\pi i/5}-e^{4\pi i /5}-e^{6\pi i /5}+e^{8 \pi i /5}$$



> 
> **Question.** Is this always possible? Stated more succinctly, can every square root be expressed as a linear combination of roots of
> unity?
> 
> 
> 



Can every square root be represented as a linear combination on roots of unity?
Yes, this is always the case.


If $\alpha=\sqrt{n}$, then $|\mathrm{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})|=2$ (since there are no intermediate fields) so $\mathrm{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}$ since that is the only group with two elements, thus it is abelian.


By the Kronecker-Weber theorem, that means that $\mathbb{Q}(\alpha)$ is contained in a cyclotomic extension, and thus it can be expressed as a sum of roots of unity. Q.E.D.


