<s>[INST] Find the sum of radicals without squaring, Is that impossible? 
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$

My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring?  It seems impossible, but I still want to ask.
 [/INST] In this answer I tried to generalize the answer given above.
Generalization:
Let,
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n,\,\,\,m\ge n$$
and
$$A=\sqrt {a+\sqrt b}+\sqrt {a-\sqrt b}=2\sqrt m$$
Then we have
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n\\
\begin{cases} m+n=a\\mn =\frac {b}{4}\end{cases}\\
t^2-at+\frac b4=0\\
4t^2-4at+b=0\\
m=\frac {a+\sqrt{a^2-b}}{2}\\
n=\frac {a-\sqrt{a^2-b}}{2}$$
This gives,
$$\sqrt {a\pm\sqrt b}=\sqrt{\frac {a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac {a-\sqrt{a^2-b}}{2}}$$
and
\begin{aligned}A&=2\sqrt m=\sqrt {4m}\\
&=\boxed{\sqrt {2\left(a+\sqrt {a^2-b}\right)}}.\end{aligned}
$\text {QED}.$
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