Convert from Nested Square Roots to Sum of Square Roots I am looking for a way to easily discover how to go from a nested root to a sum of roots.  For example, 
$$\sqrt{10-2\sqrt{21}}=\sqrt{3}-\sqrt{7}$$
I know that if I set $\alpha=\sqrt{10-2\sqrt{21}}$, square both sides, I get
$$\alpha^2=10-2\sqrt{21}$$
Now I recognize that we have a situation where $10=3+7$ and $21=7\cdot 3$, so I can immediately see that we have
$$\alpha^2=10-2\sqrt{21}=3-2\sqrt{21}+7=\sqrt{3}^2-2\sqrt{3}\sqrt{7}+\sqrt{7}^2=(\sqrt{3}-\sqrt{7})^2$$
My question is, it this the only way to approach this problem?  This approach mirrors basic algebra 1 methods of factoring quadratics, but I was curious to know if thre are other techniques that can be used to quickly deduce that a nested radical can be simplified to the sum of two radicals.  Mathematically, suppose $a,b,c,m,n,r,s\in\mathbb{N}$.  Is there a way to quickly determine $m,n,r,s$ in the equation
$$\sqrt{a\pm b\sqrt{c}}=m\sqrt{r}\pm n\sqrt{s}$$
 A: Suppose $a+b\sqrt{c}=(m\sqrt{r}+n\sqrt{s})^2$, where $a,b,c,m,n,r,s\in\mathbb{Z}$ and $r,s,c>0$. Then you have the following: $$a=m^2r+n^2s\\ b=2mn\\ c=rs\\ a+b\sqrt{c}>0$$
This shows that $b$ must be even, and the gcd's of $m,n$ and $r,s$ each divide $a$. I see no reason why this set of diophantine equations has a particularly clean solution. As an example of a case where you cannot get the desired form we have $\sqrt{64+35\sqrt{3}}$.
A: Let $a$ and $b$ be non-negative rational numbers such that
$$\sqrt{10 - 2\sqrt{21}} = \sqrt{a} - \sqrt{b}$$
Squaring both sides of the equation yields 
$$10 - 2\sqrt{21} = a - 2\sqrt{ab} + b$$
Matching rational and irrational parts yields the system of equations
\begin{align*}
a + b & = 10 \tag{1}\\
-2\sqrt{ab} & = -2\sqrt{21} \tag{2}
\end{align*}
Dividing both sides of equation 2 by $-2$ and squaring yields
$$ab = 21$$
Solving for $b$ yields
$$b = \frac{21}{a}$$
Substituting this expression in equation 1 yields
$$a + \frac{21}{a} = 10$$
Multiplying both sides of the equation by $a$ and solving the resulting quadratic equation yields
\begin{align*}
a^2 + 21 & = 10a\\
a^2 - 10a + 21 & = 0\\
(a - 3)(a - 7) & = 0
\end{align*}
The roots are $a = 3$ and $a = 7$.  If $a = 3$, then $b = 21/a = 21/3 = 7$.  However, $$\sqrt{3} - \sqrt{7} < 0 \implies \sqrt{10 - 2\sqrt{21}} \neq \sqrt{3} - \sqrt{7}$$
since the principal square root of a number cannot be negative.  
If $a = 7$, then $b = 21/a = 21/7 = 3$.  Hence, 
$$\sqrt{10 - 2\sqrt{21}} = \sqrt{7} - \sqrt{3}$$
If you instead had to evaluate $\sqrt{7 + 4\sqrt{3}}$, set 
$$\sqrt{7 + 4\sqrt{3}} = \sqrt{a} + \sqrt{b}$$
