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Following this answer, is there a simple rule for determining when: $$\sqrt{n +m\sqrt{k}}$$ Where $n,m,k \in \mathbb{N}$, can be expressed as: $$a + b\sqrt{k}$$ For some natural $a,b$?

This boils down to asking for what $n,m,k \in \mathbb{N}$ there exist $a,b\in \mathbb{N}$ such that: $$2ab= m,\ \ \text{and}\ \ a^2+b^2k = n$$

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Nathaniel Bubis
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1 Answers1

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Hint: Please see this link. To translate it to your problem, note that $ b = m^2k $. From here, one of the square roots on the right hand side of their expression must reduce.

In the case that $ a + \sqrt{a^2 - b} = 2p^2, p \in \mathbb{N} $, $ p^2 = \frac{a + \sqrt{a^2 - b}}{2} $. For the second case $ q^2 = \frac{a - \sqrt{a^2 - b}}{2} $. Hence, if $ \frac{a \pm \sqrt{a^2 - b}}{2} $ is a perfect square then the nested square root is reducible.

For example, let $ a = 30, b = 896 $. In this case, $ \frac{a + \sqrt{a^2 - b}}{2} = 16 $ so $ \sqrt{30 + \sqrt{896}} $ is reducible, being equal to $ 4 + \sqrt{14} $. Also, note that if $ a = 30, b = 756 $, it is reducible because $ \frac{a + \sqrt{a^2 - b}}{2} = 9 $, being equal to $ 3 + \sqrt{21} $.

Jon Claus
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  • I'm not sure I understood your meaning. – Nathaniel Bubis May 17 '13 at 01:00
  • $ \sqrt{a + m\sqrt{k}} = \sqrt{a + \sqrt{m^2k}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} + \sqrt{\frac{a - \sqrt{a^2 - b}}{2}} $. So it must be the case that $ a + \sqrt{a^2 - b} = 2p^2 $ or $ a - \sqrt{a^2 - b} = 2q^2 $. Then see the edit to my post. – Jon Claus May 17 '13 at 02:05