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?
 A: You're probably missing a $+$

The key is to convert it to the form $a^2+2ab+b^2$. Here you have $\sqrt{21}$ inside the outer square root, so first multiply and divide by $\sqrt 2$. Now $a$ and $b$ can be found easily as $7+3 = 10$ and $7\cdot 3 = 21$

$\begin{align}\sqrt{5+\sqrt{21}}& = \frac1{\sqrt2}\sqrt{10+2\sqrt{21}} = \frac1{\sqrt2}\sqrt{(7+3)+2\sqrt{7\cdot3}}\\& = \frac1{\sqrt2}\sqrt{(\sqrt7+\sqrt3)^2} = \frac1{\sqrt2}(\sqrt 7+\sqrt 3)\\& =\frac1{\sqrt2\cdot\sqrt2}(\sqrt {14}+\sqrt 6)\\&=\frac1{2}(\sqrt 6+\sqrt {14})\end{align}$
A: Alternative approach:
If $a,b > 0$ and $a^2 = b^2$, then $a$ must $= b$.
Let $a = \frac{1}{2} \left(\sqrt{6} + \sqrt{14}\right),
~b = \sqrt{5 + \sqrt{21}} \implies 0 < a,b.$
Then $a^2 = \frac{1}{4} \left(6 + 14 + 2\sqrt{84}\right)
= 5 + \sqrt{21} = b^2.$
A: You've requested a Galois-theoretic approach. I'm skeptical that it exists, depending on exactly what you mean; here's why.
You can use your favorite technique to show that the minimal polynomial of $\sqrt{5 + \sqrt{21}}$ and $\frac{1}{2} (\sqrt{6} + \sqrt{14})$ is $4 - 10 x^2 + x^4$. For instance, you can check these are roots and that the polynomial is irreducible. Great, that says that they're equal up to Galois conjugation. Here you can check that the Galois extension is $\mathbb{Q}(\sqrt{6}, \sqrt{14})$ and that Galois conjugation is induced by $\sqrt{6} \mapsto \pm \sqrt{6}$ and $\sqrt{14} \mapsto \pm \sqrt{14}$.
Ok, but are they exactly equal, or are they merely conjugates? You could have picked $\sqrt{21} = \color{red}{-}4.58...$ and positives for all the other roots, in which case they actually wouldn't be equal. So, whether they're equal depends on your choice of interpretation of $\sqrt{21}$. The Galois-theoretic considerations above are all independent of this choice, but your literal question is not.
A: 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, $\alpha = \frac12[(\alpha+\beta)+(\alpha-\beta)] = \frac12(\sqrt6+\sqrt{14})$.
