<s>[INST] Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\frac{1}{\sqrt{2}}\left(\sqrt{46-6\sqrt{5}}-2\sqrt{6-2\sqrt{5}}\right)=$$
$$=\frac{1}{\sqrt{2}}\left(\sqrt{(3\sqrt{5}-1)^2}-2\sqrt{(\sqrt5-1)^2}\right)=\frac{1}{\sqrt{2}}\left(3\sqrt{5}-1-2(\sqrt{5}-1)\right)=$$
$$=\frac{1}{\sqrt{2}}(\sqrt{5}+1)=\frac{1}{\sqrt{2}}\sqrt{6+2\sqrt{5}}=\sqrt{3+\sqrt{5}}.$$
But this way does not work for the starting identity. 
How to prove the starting identity without squaring? 
Thank you!  
 [/INST] It still works, albeit cumbersome 
\begin{align}
& \sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}\\
= & \frac12 \sqrt{\left(\sqrt{1775-425\sqrt{17}} +\sqrt{1207-289\sqrt{17}} \right)^2}
-\frac12 \sqrt{\left(\sqrt{639-153\sqrt{17}} +\sqrt{1207-289\sqrt{17}} \right)^2}\\
= & \frac52 \sqrt{71-17\sqrt{17}} +\frac12\sqrt{1207-289\sqrt{17}}
-\frac32 \sqrt{71-17\sqrt{17}} -\frac12 \sqrt{1207-289\sqrt{17}} \\
=&\left( \frac52-\frac32\right) \sqrt{71-17\sqrt{17}}\\
=& \sqrt{71-17\sqrt{17}}\\
\end{align}
</s>