<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] A polynomial approach.
Note that $a_1,a_2=\sqrt{3\pm\sqrt 5}$ are two of the roots of $x^4-6x^2+4=0.$ The other roots are $-a_1,-a_2.$
So $$x^4-6x^2+4=(x-a_1)(x-a_2)(x+a_1)(x+a_2).$$
Now, $a_1a_2=\sqrt{4}=2.$ If $S=a_1+a_2,$ then this factorization becomes:
$$x^4-6x^2+4=(x^2-Sx+2)(x^2+Sx+2)=x^4+(4-S^2)x^2+4.$$
So $4-S^2=-6,$ or $S^2=10.$
I guess that's sort of squaring, but we never actually numerically square $S.$
More generally, given the roots of $x^4-bx^2+c^2,$ there are two roots $a_1,a_2$ with $a_1a_2=c,$ and then you get a similar result:
$$x^4-bx^2+c^2=(x^2-Sx+c)(x^2+Sx+c)=x^4+(2c-S^2)x^2+c^2,$$ so $2c+b=S^2.$
So this means, at least if $c\geq 0,$ that $$\sqrt{\frac{b+\sqrt{b^2-4c^2}}2}+\sqrt{\frac{b-\sqrt{b^2-4c^2}}2}=\pm\sqrt{2c+b}$$
Multiplying by $\sqrt{2},$ this gives:
$$\sqrt{b+\sqrt{b^2-4c^2}}+\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{4c+2b}\tag1$$
You also get:
$$\sqrt{b+\sqrt{b^2-4c^2}}-\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{2b-4c},\tag 2$$ since in this case $a_1a_2=-c.$
The case $b=3,c=1$ in (1) gives your result, since $3\pm\sqrt 5$ are both real and positive, so we know the result has to be positive.
In both $(1)$ and $(2)$ you get the positive sign if $b,c$ and all the square roots are real. If the square roots are complex, you are stuck figuring out the sign.
But if $b,c$ are real, and $0\leq 2c\leq b,$ we can solve $(1)$ and $(2)$ to get:
$$\sqrt{b+\sqrt{b^2-4c^2}}=\frac{\sqrt{2b+4c}+\sqrt{2b-4c}}{2}=\frac{\sqrt{b+2c}+\sqrt{b-2c}}{\sqrt 2}$$
and similarly:
$$\sqrt{b-\sqrt{b^2-4c^2}}=\frac{\sqrt{b+2c}-\sqrt{b-2c}}{\sqrt 2}$$
If $d=b^2-4c^2,$ this gives us:
$$\sqrt{b+\sqrt{d}}=\frac{\sqrt{b+\sqrt{b^2-d}}+\sqrt{b-\sqrt{b^2-d}}}{\sqrt2}$$ when $0\leq d\leq b^2.$
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