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Suppose we are given a quadratic number field $\mathbb{Q}(\sqrt{d})$, for some integer $d$ which is not a perfect square. I wish to study when is an element $\alpha \in \mathbb{Q}(\sqrt{d})$ a perfect square, i.e. $\alpha = \beta^2$ for some $\beta \in \mathbb{Q}(\sqrt{d})$. Similarly, how to test whether an element $\alpha \in \mathbb{Q}(\sqrt{d})$ is a perfect cube, i.e. $\alpha = \beta^3$ for some $\beta \in \mathbb{Q}(\sqrt{d})$. I know the above questions reduce to solving appropriate quadratic/cubic equations over rationals but I am looking for a direct test similar to quadratic/cubic residuacity tests that we have for finite fields.

I have little to zero knowledge of algebraic number theory but I am willing to do the hard work. I am looking for books or resources (preferably beginner friendly) which deal with these two questions in detail. My gratitude in advance if anyone can point me in the right direction.

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    $\begingroup$ Do you mean squares or quadratic residues? $\endgroup$ Jun 24, 2021 at 12:03
  • $\begingroup$ @franzlemmermeyer I guess square will be the appropriate term. $\endgroup$ Jun 24, 2021 at 17:53
  • $\begingroup$ @franzlemmermeyer I realized just now that I had looked up your website in fen.bilkent.edu.tr/~franz/rl2/rlb12.pdf before asking this question. Based on the little knowledge that I have, I am aware that quadratic and cubic reciprocity laws do go into $\mathbb{Z}[\iota]$ and $\mathbb{Z}[\omega]$ but I am not sure whether the quadratic reciprocity in number fields in the link above actually has any connection to my query above, as you correctly pointed out that I am looking for perfect squares and cubes, and not residues in number fields. Could you please clarify that, if possible. $\endgroup$ Jun 25, 2021 at 11:09

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Since every algebraic number is the quotient of an algebraic integer and an ordinary integer, we can write every element as a quotient of an algebraic integer and a square number. This reduced the problem to elements of ${\mathbb Z}[\sqrt{m}]$.

A necessary condition for $\alpha$ to be a square is that its norm is a square; assume therefore that $N\alpha = \alpha \alpha' = m^2$ for an integer $n$.

Now we recall the ancient Babylonians' idea that you can compute two unknowns from their sum and their difference: \begin{align*} (\sqrt{\alpha} + \sqrt{\alpha'}\,)^2 & = \alpha + \alpha' + 2n, \\ (\sqrt{\alpha} - \sqrt{\alpha'}\,)^2 & = \alpha + \alpha' - 2n, \end{align*} Thus with $r = \sqrt{\alpha + \alpha' + 2n}$ and $s = \sqrt{\frac{\alpha + \alpha' - 2n}m}$ (if $s$ is not an integer, $\alpha$ is not a square) we have $$ \sqrt{\alpha} = \frac{r + s \sqrt{m}}2. $$

Example: Compute $\sqrt{\alpha}$ for $\alpha = 37 + 20 \sqrt{3}$. Here $N\alpha = 169 = 13^2$, and we find $r = \sqrt{74 + 2 \cdot 13} = 10$, $s = \sqrt{\frac{74 - 2\cdot 13}3} = 4$, hence $\sqrt{\alpha} = 5 + 2 \sqrt{3}$.

Of course you can test whether elements are squares by testing whether they are squares modulo suitably chosen ideals; if you choose to do so, this is best done without using any reciprocity law.

A necessary condition for $\alpha = r + s \sqrt{m}$ to be a cube is that its norm $\alpha \alpha' = n^3$ is the cube of an integer. In this case $$ (\sqrt[3]{\alpha} + \sqrt[3]{\alpha'})^3 = \alpha + \alpha' + 3n (\sqrt[3]{\alpha} + \sqrt[3]{\alpha'}). $$ For $\alpha$ to be a cube, this equation must have an integral root $2r$, which can be easily checked in a finite number of steps.

Next consider $\omega = \frac{\sqrt[3]{\alpha} - \sqrt[3]{\alpha'}}{\sqrt{m}}$; here $$ \omega^3 = \frac{\alpha - \alpha'}{m \sqrt{m}} + \frac{3n}{m} \omega, $$ and if $\alpha$ is a cube, this cubic must have an integral root $2s$.

Example: Let $\alpha = 100 + 51 \sqrt{3}$. Then $N\alpha = \alpha \alpha' = 13^3$, and $2r$ must be an integral root of $$ X^3 - 39X -200 = 0. $$ Since $X = 8$ is the only real root, we must have $2r = 8$, hence $r = 4$. Similarly, $2s$ is a root of $$ X^3 + 13X - 34 = 0, $$ and since $X = 2$ is the only real root we conclude that $s = 1$. In fact, $\sqrt[3]{100 + 51\sqrt{3}} = 4 + \sqrt{3}$.

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  • $\begingroup$ Is there a necessary condition for $\alpha$ to be a cube in $\mathbb{Z}[\sqrt{m}]$? I mean do we have something similar to test perfect cubes also? $\endgroup$ Jun 28, 2021 at 4:39
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    $\begingroup$ @Pranav The quadratic denesting formula has a simple memorable form that I describe here. $\endgroup$ Dec 8, 2021 at 15:59
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The answer depends on the kind of "test" you would hope to have. Given $\alpha\in\mathbb{Q}(\sqrt{d})$, then $\alpha=a+b\sqrt{d}$ for some $a,b\in\mathbb{Q}$. What you are asking for is: on what conditions $\sqrt{\alpha}$ is again an element of $\mathbb{Q}(\sqrt{d})$? i.e. we must find $\lambda,\mu\in\mathbb{Q}$ such that $\sqrt{\alpha}=\lambda+\mu\sqrt{d}$. By last equality, if you square both sides you get:

$\alpha=\lambda^2+\mu^2d+2\lambda\mu\sqrt{d}$. But also $\alpha=a+b\sqrt{d}$ as initially stated. Since $\{1,\sqrt{d}\}$ is a base, by: $$\lambda^2+\mu^2d+2\lambda\mu\sqrt{d}=a+b\sqrt{d}$$ it follows: $$a=\lambda^2+\mu^2d$$ and $$b=2\lambda\mu$$ The most interesting is the first one, it basically tell you when the square root of $\alpha$ does live in $\mathbb{Q}(\sqrt{d})$: indeed we wrote $a=\lambda^2+\mu^2d$, which doesn't have solutions when $a$ is negative and $d$ positive. On the other hand, suppose $d>0$ and $a>0$. By that system you reach the equation: $$4d\mu^4-4a\mu^2+b^2=0$$ and $\frac{\Delta}{2}=4a^2-4db^2$ must be $0$ in order to have a "unique solution" ($2$ instead of $4$). Thus $a^{2}=db^2$, but now we have a problem: it follows that $\sqrt{d}=\frac{a}{b}$ which is impossible unless $d$ is a perfect square. Just to trying and keep going, assume then $\frac{\Delta}{2}>0$, it would lead you to: $$\mu^2=\frac{a\pm\sqrt{a^2-db^2}}{2d}$$

Now, remember we want $\mu\in\mathbb{Q}$ and all these last passages brings us to say: $$\mu=\pm\sqrt{\frac{a\pm\sqrt{a^2-db^2}}{2d}}$$ Basically, as test you may have $a^2-db^2>0$ and $\frac{a\pm\sqrt{a^2-db^2}}{2d}$ being a perfect square.

A cubic root I suspect would require you much more computations and details.

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  • $\begingroup$ If you multiply second equation by $a$ and subtract it from first equation multiplied by $b$, you will arrive at the condition that $4a^2-4db^2$ is a perfect square $> 0$. I borrowed this idea from math.stackexchange.com/questions/550121/… which works for cubics. Unfortunately, in my research the original problem I started with was testing solvability of a cubic equation over rationals, which one can show reduces to testing whether an element in number field is a cube. Ofcourse one can directly solve it using LLL but I want easier test $\endgroup$ Jun 25, 2021 at 10:55
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In the case that $d<0$, there is even another method which is quite simple and straightforward in my mind. However: It only works for $d<0$.

So, assume $d<0$ and write $K=\mathbb{Q}(\sqrt{d})$ and $R=\mathbb{Z}[\sqrt{d}]$. As explained in franz lemmermeyer's answer, we reduce the problem to $R$. So, we assume $\alpha \in R$ and we want to check whether it is a square in $R$. Compute its norm: $$N(\alpha)=m$$ where $m \in \mathbb{Z}$. Now, every $\beta \in R$ with $\beta^2=\alpha$ must satisfy $$N(\beta)^2=m.$$ Write $\beta=x+\sqrt{d}y$ with $x,y \in \mathbb{Z}$. Plugging this into the above equation gives $$(x^2+|d|y^2)^2=m.$$ This equation has only finitely many solutions $x,y \in \mathbb{Z}$ since each solution must satify $|x| \leq \sqrt{m}$ and $|y|\leq \sqrt{m/|d|}$ and there are only finitely many integers with this property. (The bounds $\sqrt{m}$ and $\sqrt{m/|d|}$ can even be improved.) To check whether $\alpha$ is a square you have to complete the following two steps:

Step 1: Compute all integer solutions of $(x^2+|d|y^2)^2=m$. (Just use trial and error for all integers $x,y$ less than the bounds mentioned above.)

Step 2: For all solutions $x,y$ from step 1, check whether $(x+\sqrt{d}y)^2=\alpha$.

If step 2 was successful, $\alpha$ is a square. If not, $\alpha$ is not a square.

The question whether $\alpha$ is a cube is the same except that you have to deal with an equation of the form $N(\beta)^3=m$ or, more precisely, $$(x^2+|d|y^2)^3=m.$$ In fact, this method works for any power (as long as $d<0$).

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