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Let $m$ be a positive square-free integer, and a number $z\in Q(\sqrt{m})$ is given. How to find a square root of $z$, if it exists in $Q(\sqrt{m})$?

Here $Q$ is the field of rational numbers.

I am asking for a reference or an explanation of an algorithm. Of course, I can look for a square root in the form $x+y\sqrt{m}$, square this, and obtain a system of quadratic equations. But solving this system leads again to a question of the same kind. Namely, if $z=a+b\sqrt{m}$ I obtain: $$x^2+my^2=a,$$ $$2xy=b.$$ This is equivalent to a biquadratic equation $4x^4-4ax^2+mb^2=0$, and $x^2=(4a\pm\sqrt{16a^2-mb^2})/8$, so I am back to a problem of the same kind.

For example, for the $\sqrt{15/2+(5/2)\sqrt{5}}$. I can guess the answer, but I want to have an algorithm.

Alexandre Eremenko
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  • I am not understanding your question. It appears to me that using the technique you suggested, you would not be guessing at an answer, but calculating one. In the example you give using the method you give, you wind up with two possible values for the square root: $$\pm \dfrac{5}{2}\pm \dfrac{\sqrt{5}}{2}$$ and squaring both of those possibilities gives the same outcome, proving they are both valid roots. – SlipEternal Nov 05 '19 at 19:20
  • @InterstellarProbe: I added explanation. – Alexandre Eremenko Nov 05 '19 at 19:36
  • I posted as an answer since it was too long for a comment. – SlipEternal Nov 05 '19 at 19:54

1 Answers1

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You have:

$$x^2+my^2=a \\ 2xy = b$$

If instead you solve for $y$, you get:

$$4my^4 - 4ay^2+b^2 = 0$$

This distinction may not be important, but it is a bit easier to solve. Solving for $y^2$ gives:

$$y^2 = \dfrac{4a \pm \sqrt{16a^2-16mb^2}}{8m} = \dfrac{a \pm \sqrt{a^2-mb^2}}{2m}$$

But, you must have $y \in \mathbb{Q}$, so it must be that $\sqrt{a^2-mb^2} \in \mathbb{Q}$.

In the example you gave with $z = \dfrac{15}{2}+\dfrac{5}{2}\sqrt{5}$, you have:

$$y^2 = \dfrac{\tfrac{15}{2}\pm \sqrt{\tfrac{225}{4}-\tfrac{125}{4}}}{10} = \dfrac{\tfrac{15}{2}\pm \tfrac{10}{2}}{10}$$

$$y^2 = \dfrac{1}{4}\text{ or }y^2 = \dfrac{5}{4}$$

Next, you must have $y \in \mathbb{Q}$, so it must be that $y = \pm \dfrac{1}{2}$.

Now, we have

$$x = \dfrac{5}{4y} = \pm \dfrac{5}{2}$$

And this gives us our two answers:

$$\sqrt{\dfrac{15}{2}+\dfrac{5}{2}\sqrt{5}} = \pm\left( \dfrac{5}{2}+\dfrac{1}{2}\sqrt{5}\right)$$

And the requirement that $x,y \in \mathbb{Q}$ will always limit your answers so that you are no longer "guessing".

SlipEternal
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