Playing with Maple, I noticed that it gives the square root of $c = 1+\frac{\sqrt3}{2}$ as equal to $a = \frac{1}{2}+\frac{\sqrt3}{2}$.

Indeed it checks out. But I got curious: how can I find that value, or more generally any square root of numbers of the form $x+y\sqrt{k}$?

I was able to do it the following way: the square of $z = x+y\sqrt{k}$ is also of the same form. Therefore, I can suppose there is a number of that form whose square is equal to $z$.

In my case, I want to find $(x,y)$ such that $(x+y\sqrt3)^2 = 1+\frac{\sqrt3}{2}$. I developed, which yields $(x^2+3y^2) + 2xy\sqrt3 = 1+\frac{\sqrt3}{2}$. Then I matched the coefficients of $1$ and of $\sqrt3$ on both sides, to get the system:

$x^2+3y^2 = 1$

$2xy = \frac{1}{2}$

Solving for $x$ and $y$, I got $x = ±\frac{1}{2}$ and $y = ±\frac{1}{2}$ (there is another pair of solutions that compute to the same number). QED.

Is my method correct? Is there any more efficient way? Is it possible to prove that a solution of the form $z = x+y\sqrt{k}$ always exist and if not, when?



3 Answers 3


Simpler than undetermined coefficients is the following rule I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here $\:1+\sqrt{3}/2\:$ has norm $= 1/4.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 1/2\ $ yields $\ 1/2+\sqrt{3}/2\:$

and this has $\rm\ \sqrt{trace}\: =\: 1,\ \ thus,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1/2+\sqrt{3}/2.$

Below is another example.

Note $\:9-4\sqrt{2}\:$ has norm $= 49.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 7\ $ yields $\ 2-4\sqrt{2}\:$

and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ so,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1-2\sqrt{2}.$

See here for many more examples, and see this answer for general radical denesting algorithms.

  • $\begingroup$ I checked your answer as the correct one because I like the trick for its fun and relative simplicity. $\endgroup$ Sep 16, 2012 at 7:41

For rationals $x,y,a,b,k$ where $k$ is square-free and $y \ne 0$, $(a + b \sqrt{k})^2 = x + y \sqrt{k}$ if and only if $a^2 + k b^2 = x$ and $2 a b = y$, and thus $b = y/(2a)$ and $a^2 + k y^2/(4 a^2) = x$, i.e. $(2 a^2 - x)^2 = x^2 - k y^2$. Thus $x^2 - k y^2$ must be the square of a rational, and if it is $r^2$ then $(x \pm r)/2$ must be the square of a rational.

In your example, $x=1, y=1/2, k=3$, $x^2 - k y^2 = 1 - 3/4 = 1/4 = (1/2)^2$ and $(1 - 1/2)/2 = (1/2)^2$, so $1 + \sqrt{3}/2$ is a square.

On the other hand, for $x=3, y=1, k=5$, $x^2 - k y^2 = 9 - 5 = 2^2$ but $(3 - 2)/2 = 1/2$ and $(3+2)/2 = 5/2$ are not squares, so $3 + \sqrt{5}$ is not the square of an expression of the form $a + b \sqrt{5}$ with $a$ and $b$ rational (on the other hand, it is the square of $\sqrt{10}/2 + \sqrt{2}/2$).

  • $\begingroup$ Thanks, though my question was not about rational coefficients specifically. I was more concerned about the validity of my therefore, for which your answer is useful. $\endgroup$ Sep 16, 2012 at 7:40

The method is correct. Unfortunately, the square roots of $s+t\sqrt{k}$, where $s$ $t$ and $k$ are rational, are usually not of the form $x+y\sqrt{k}$ with rational $x$ and $y$. When you use your method, the problem gets revealed when the system of equations analogous to the one you produced has no rational solution.

The system of equations can be turned into a quadratic in $x^2$. If the discriminant is not the square of a rational, there are no rational $x$ and $y$. And even if the discriminant is the square of a rational, giving rational $x^2$, that does not guarantee the rationality of $x$.

  • $\begingroup$ Thanks. My question was not particularly about rational coefficient, but the consideration is interesting. $\endgroup$ Sep 16, 2012 at 7:38

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