Convert from Nested Square Roots to Sum of Square Roots

Question

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

$$\sqrt{a\pm b\sqrt{c}}=m\sqrt{r}\pm n\sqrt{s}$$

You are able to get a nested radical rewritten as a $\mathbb{Q}$-linear combination of square roots because you are working in an extension of $\mathbb{Q}$ that is the splitting field of a polynomial that is the product of two irreducible quadratics. — J. David Taylor, May 08 '16 at 02:55
This seems relevant, if you drop the restriction to integers: http://math.stackexchange.com/questions/660161/how-to-arrive-at-ramanujan-nested-radical-identity?rq=1 — J. David Taylor, May 08 '16 at 04:31
See also: Denesting radicals at Wikipedia, Strategies to denest nested radicals (and the posts linked there), Denesting a square root: $\sqrt{7 + \sqrt{14}}$ (and the posts linked there) — Martin Sleziak, Oct 16 '16 at 09:15

score 3 · Accepted Answer · answered May 08 '16 at 02:50

3

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 $\sqrt{64+35\sqrt{3}}$.

answered May 08 '16 at 02:50

J. David Taylor

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What I thought. Rearranging with a formula just doesn't work. It's like finding the formula of a number's HCF. – Bradman175 May 08 '16 at 03:37
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Yeah, if I had a formula for that, I'd be rich. – J. David Taylor May 08 '16 at 04:27
you could make a program to solve it lol. – Bradman175 May 08 '16 at 04:28
Oh, yeah. I suppose the Euclidean algorithm does the trick. I somehow read your comment as implying an efficient algorithm for factoring integers. hehe – J. David Taylor May 08 '16 at 04:29

score 3 · Answer 2 · answered May 08 '16 at 11:27

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 $$\sqrt{7 + 4\sqrt{3}} = \sqrt{a} + \sqrt{b}$$

Convert from Nested Square Roots to Sum of Square Roots

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