When is $\sqrt{n-\sqrt{m}}=\sqrt{a}-\sqrt{b}$ where $m,n\in\Bbb{Z}^+$ and $a,b\in\Bbb{Q}$?

Asked Dec 20 '22 at 15:55

Active Dec 20 '22 at 18:17

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In this youtube video, Pratik Matematik asked "$\sqrt{\sqrt{49}-\sqrt{48}}=?$".

I generalized it as $\sqrt{n-\sqrt{n^2-k^2}}=\sqrt{\frac{n+k}{2}}-\sqrt{\frac{n-k}{2}}$.

And the solution of PM's question is $\sqrt{\frac{7+1}{2}}-\sqrt{\frac{7-1}{2}}=2-\sqrt{3}.$

Can you give another family of examples where $m$ is not in the form $n^2-k^2$ and still $\sqrt{n-\sqrt{m}}=\sqrt{a}-\sqrt{b}$ where $n,m$ are positive integers and $a,b\in\Bbb{Q}$?

Thanks in advance!

asked Dec 20 '22 at 15:55

Bob Dobbs

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Squaring both sides gives $n=a+b, m=4ab.$ But $4ab=n^2-(a-b)^2.$ So if $k=a-b,$ it seems like you will get your general case. – Thomas Andrews Dec 20 '22 at 16:08
So the general cases will probably be cases where $\sqrt m$ is rational, which are not interesting, but certainly can be made an infinite family. – Thomas Andrews Dec 20 '22 at 16:09
@ThomasAndrews So the question is not interesting? – Bob Dobbs Dec 20 '22 at 16:17
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Well, the question is interesting, but the answer is that there are no interesting examples. – Thomas Andrews Dec 20 '22 at 16:31
Your generalization is a special case of the general formula I give here in the linked dupe (which always works when a denesting exists). In the proof there we have $,s = n+k-\sqrt{n^2-k^2},,$ and $,t = \sqrt{2(n+k)},,$ and the result $,s/t,$ yields your formula. – Bill Dubuque Dec 20 '22 at 18:14
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More generally see the Radical Denesting Structure Theorem and the citations there for general theory. – Bill Dubuque Dec 20 '22 at 18:18

When is $\sqrt{n-\sqrt{m}}=\sqrt{a}-\sqrt{b}$ where $m,n\in\Bbb{Z}^+$ and $a,b\in\Bbb{Q}$?

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