<s>[INST] Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?
 [/INST] A Galois-theoretic approach would analyse the conjugates. We know that the conjugates of $\alpha := \sqrt{5+\sqrt{21}}$ are $\pm\sqrt{5 \pm \sqrt{21}}$ and the conjugates of $\frac12(\sqrt6 + \sqrt{14})$ are $\frac12(\pm \sqrt6 \pm \sqrt{14})$, so the interesting conjugate to analyse would be $\beta := \sqrt{5-\sqrt{21}}$ (the other two are $-\alpha$ and $-\beta$).
We compute:
$$\begin{array}{rcl}
(\alpha + \beta)^2
&=& \alpha^2 + \beta^2 + 2\alpha\beta \\
&=& (5+\sqrt{21}) + (5-\sqrt{21}) + 2\sqrt{(5+\sqrt{21})(5-\sqrt{21})} \\
&=& 10 + 2\sqrt{5^2 - 21} \\
&=& 14
\end{array}$$
So $\alpha + \beta = \sqrt{14}$ because it is positive.
We also compute:
$$\begin{array}{rcl}
(\alpha-\beta)^2
&=& \alpha^2+\beta^2-2\alpha\beta \\
&=& 10 - 2\sqrt{5^2-21} \\
&=& 6
\end{array}$$
So $\alpha - \beta = \sqrt6$ because $\alpha > \beta$.
Therefore, $\alpha = \frac12[(\alpha+\beta)+(\alpha-\beta)] = \frac12(\sqrt6+\sqrt{14})$.
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