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By nested radical, I mean an expression of the form $\sqrt{a+b\sqrt{n}}$ where a, b and n are positive integers and n is not a perfect square.

I wrote a computer program that randomly generated pairs of nested radicals (with a common value of n) such that neither nested radical could denest but their product could. One example is:

$$\sqrt{8+2\sqrt{11}} \times \sqrt{13+2\sqrt{11}} = 7+3\sqrt{11}$$

But when I randomly generated pairs of nested radicals (with a common value of n) such that neither nested radical could denest but hoping their sum could, I couldn’t find any. Do any such pairs exist? If so, I’d like a few examples and some guidance about how to find more. If such pairs don’t exist, I’d like to know why not.

Will Octagon Gibson
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1 Answers1

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It is not possible. Suppose the following sum denested:

$$ \sqrt{a+b\sqrt{n}}+\sqrt{c+d\sqrt{n}}= u + v \sqrt{n} \tag{1} $$

Then the difference of the two radicals on the LHS would also denest (assuming $\,u,v\,$ rational and $\,\sqrt{n}\,$ irrational, so $\,u^2 - v^2n \ne 0\,$):

$$ \begin{align} \sqrt{a+b\sqrt{n}} - \sqrt{c+d\sqrt{n}} &= \frac{\left(a+b\sqrt{n}\right)-\left(c+d\sqrt{n}\right)}{\sqrt{a+b\sqrt{n}}+\sqrt{c+d\sqrt{n}}} \\ &= \frac{(a-c) + (b-d)\sqrt{n}}{u+v\sqrt{n}} \color{blue}{\cdot \frac{u - v\sqrt{n}}{u-v\sqrt{n}}} \\ &= \frac{(a-c)u-(b-d)vn + \left((b-d)u-(a-c)v\right)\sqrt{n}}{u^2 - v^2n} \\ &= u' + v' \sqrt{n} \tag{2} \end{align} $$

Adding $\,(1)+(2)\,$ gives $\,\sqrt{a+b\sqrt{n}} = \dfrac{1}{2}\big((u+u') + (v+v')\sqrt{n}\big)\,$, and subtracting $\,(1)-(2)\,$ gives $\,\sqrt{c+d\sqrt{n}} = \dfrac{1}{2}\big((u-u') + (v-v')\sqrt{n}\big)\,$, so both radicals denest.

dxiv
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  • Why would the sum denesting imply that the difference would also denest? Also why is $u^2-v^2n$ not equal to 0? – Will Octagon Gibson Jun 06 '22 at 06:44
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    @WillOctagonGibson $;$ 1) The sequence of equalities ending with $(2)$ proves that the difference does also denest. The second '$=$' sign makes use of the premise that the sum denests, and the rest is just straightforward algebraic manipulations showing that the difference can be written in the form $u'+v'\sqrt{n}$, meaning the difference also denests. $;$ 2) Because $u^2-v^2n=0$ would imply that $\sqrt{n} = \frac{|u|}{|v|}$ is rational, contrary to the assumption that it's not (from your question: "$n$ is not a perfect square"). – dxiv Jun 06 '22 at 07:22
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    @WillOctagonGibson The answer not only proves that both radicals do in fact denest, but also gives you the steps to determine the actual denesting. You can try it on a concrete case to see how it all works out. – dxiv Jun 06 '22 at 07:34
  • I understand your 1st comment but not your 2nd comment. What sort of concrete case are you talking about considering that the sums can’t denest? – Will Octagon Gibson Jun 06 '22 at 09:19
  • @WillOctagonGibson Suppose you are given that $,\sqrt{{3} - 2 \sqrt{2}} + \sqrt{22 + 12 \sqrt{2}} = 1 + 4 \sqrt{2},$. Then the steps above show that $,\sqrt{{3} - 2 \sqrt{2}} = \sqrt{2} -1,$ and $,\sqrt{22 + 12 \sqrt{2}} = 2 + 3 \sqrt{2},$, which proves that both radicals on the LHS of the original equality do in fact denest, and gives the explicit denesting for each. – dxiv Jun 06 '22 at 16:06
  • I understand your concrete case comment. – Will Octagon Gibson Jun 06 '22 at 18:22
  • Your proof by contradiction uses the idea that the sum (if it could denest) would denest in the form $u+v\sqrt{n}$. Is it possible that the sum could denest in some other form? – Will Octagon Gibson Jun 06 '22 at 18:31
  • @WillOctagonGibson Not sure what other form of denesting you have in mind. If you think at something like $,u+v\sqrt{n}+w\sqrt{m},$, then that's not possible (except for the trivial case when $m/n$ is a perfect rational square, so $,\sqrt{n},$ and $,\sqrt{m},$ are rational multiples of one another). This ultimately follows from the linear independence of "unrelated" irrational square roots over the rationals, see for example 1, 2, 3. – dxiv Jun 06 '22 at 20:20