Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?

Question

In this recent answer to this question by Eesu, Vladimir Reshetnikov proved that $$ \begin{equation} \left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1} \end{equation} $$

I would like to know if this result can be generalized to other triples of natural numbers.

Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left( p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right) \in \mathbb{N} ^{3}.\tag{2} \end{equation} $$

For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$

$$ 26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \sqrt{3} $$

and solve the system $$ \left\{ \begin{array}{c} a^{3}+9ab^{2}=26 \\ a^{2}b+b^{3}=5. \end{array} \right. $$

A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.

For $(2)$ the very same idea yields

$$ p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3} \sqrt{3} $$

and

$$ \left\{ \begin{array}{c} a^{3}+9ab^{2}=p \\ 3( a^{2}b+b^{3}) =q. \end{array} \right. \tag{4} $$

I tried to solve this system for $a,b$ but since the solution is of the form

$$ (a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5} $$

where $x$ satisfies the cubic equation $$ 64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6} $$ would be very difficult to succeed, using this naive approach.

Is this problem solvable, at least partially?

Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?

Answer 1

The solutions are of the form $\displaystyle(p, q)= \left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right)$, for any rational parameter $t$. To prove it, we start with $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n\tag{$\left(p,q,n,r\right)\in\mathbb{N}^{4}$}$$ and cube both sides using the identity $(a+b)^3=a^3+3ab(a+b)+b^3$ to, then, get $$\left(\frac{n^3-2p}{3n}\right)^3=p^2-rq^2,$$ which is a nicer form to work with. Keeping $n$ and $r$ fixed, we see that for every $p={1,2,3,\ldots}$ there is a solution $(p,q)$, where $\displaystyle q^2=\frac{1}{r}\left(p^2-\left(\frac{n^3-2p}{3n}\right)^3\right)$. When is this number a perfect square? Wolfram says it equals $$q^2 =\frac{(8p-n^3) (n^3+p)^2}{(3n)^2\cdot 3nr},$$ which reduces the question to when $\displaystyle \frac{8p-n^3}{3nr}$ is a perfect square, and you get solutions of the form $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right).$ Note that when $r=3$, this simplifies further to when $\displaystyle \frac{8p}{n}-n^2$ is a perfect square.

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Now, we note that if $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right) \in\mathbb{Q}^2$, $\displaystyle\sqrt{\frac{8p-n^3}{3nr}}$ must be rational as well. Call this rational number $t$, our parameter. Then $8p=3t^2nr+n^3$. Substitute back to get $$(p,q)=\left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right).$$ This generates expressions like $$\left(\frac{2589437}{8}+\frac{56351}{4}\sqrt{11}\right)^{1/3}+\left(\frac{2589437}{8}-\frac{56351}{4}\sqrt{11}\right)^{1/3}=137$$

$$\left(\frac{11155}{4}+\frac{6069}{4}\sqrt{3}\right)^{1/3}+\left(\frac{11155}{4}-\frac{6069}{4}\sqrt{3}\right)^{1/3}=23$$

for whichever $r$ you want, the first using $(r,t,n)=(11,2,137)$ and the second $(r,t,n)=(3,7,23)$.

Answer 2

Using the fact that: $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ and letting $x=m^{\frac{1}{3}}$ and $y=n^{\frac{1}{3}}$, we get the statement: $$m+n=(m^{\frac{1}{3}}+n^{\frac{1}{3}})(m^\frac{2}{3}-(mn)^{\frac{1}{3}}+n^{\frac{2}{3}})$$Maybe this will help.

Answer 3

There is an infinite family of solutions coming from the idea $(2+\sqrt 3)^3=26+15\sqrt 3$. We can form $(2+\sqrt 3)^{3n}$ and find another solution. The next one is $(2+\sqrt 3)^6=1351+780 \sqrt 3$ and $(1351+780\sqrt 3)^{(1/3)}+(1351-780\sqrt 3)^{(1/3)}=14$ There is a recurrence, if $(a,b)$ is a solution, the next is $(26a+45b,15a+26b)$ and so we get triplets $(1351,780,14),(70226,40545,52),(3650401,2107560,194),(189750626,109552575,724) 9863382151,5694626340,2702)$

and on. I have not shown that these are all the solutions.

Answer 4

Here's a way of finding, at the very least, a large class of rational solutions. It seems plausible to me that these are all the rational solutions, but I don't actually have a proof yet...

Say we want to solve $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$ for some fixed $n$. The left-hand side looks an awful lot like the root of a depressed cubic (as it would be given by Cardano's formula). So let's try to build some specific depressed cubic having $n$ as a root, where the cubic formula realizes $n$ as $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$.

The depressed cubics having $n$ as a root all take the following form: $$(x-n)(x^2+nx+b) = x^3 + (b-n^2)x-nb$$ where $b$ is arbitrary. If we want to apply the cubic formula to such a polynomial and come up with the root $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$, we must have: \begin{eqnarray} p&=&\frac{nb}{2}\\ 3q^2&=& \frac{(nb)^2}{4}+\frac{(-n^2+b)^3}{27}\\ &=&\frac{b^3}{27}+\frac{5b^2n^2}{36}+\frac{bn^4}{9}-\frac{n^6}{27}\\ &=&\frac{1}{108}(4b-n^2)(b+2n^2)^2 \end{eqnarray} (where I cheated and used Wolfram Alpha to do the last factorization :)).

So the $p$ that arises here will be rational iff $b$ is; the $q$ that arises will be rational iff $4b-n^2$ is a perfect rational square (since $3 * 108=324$ is a perfect square). That is, we can choose rational $n$ and $m$ and set $m^2=4b-n^2$, and then we will be able to find rational $p,q$ via the above formulae, where $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$ is a root of the cubic $$ (x-n)\left(x^2+nx+\frac{m^2+n^2}{4}\right)=(x-n)\left(\left(x+\frac{n}{2}\right)^2+\left(\frac{m}{2}\right)^2\right) \, . $$

The quadratic factor of this cubic manifestly does not have real roots unless $m=0$; since $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}$ is real, it must therefore be equal to $n$ whenever $m \neq 0$.

To summarize, we have found a two-parameter family of rational solutions to the general equation $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$. One of those parameters is $n$ itself; if the other is $m$, we can substitute $b=\frac{m^2+n^2}{4}$ into the above relations to get \begin{eqnarray} p&=&n\left(\frac{m^2+n^2}{8}\right)\\ q&=&m\left(\frac{m^2+9n^2}{72}\right) \, . \end{eqnarray}

To make sure I didn't make any algebra errors, I randomly picked $n=5$, $m=27$ to try out. These give $(p,q)=\left(\frac{1885}{4},\frac{1431}{4}\right)$, and indeed Wolfram Alpha confirms that $$ \left(\frac{1885}{4}+\frac{1431}{4} \sqrt{3}\right)^{1/3}+\left(\frac{1885}{4}-\frac{1431}{4} \sqrt{3}\right)^{1/3}=5 \, . $$

Answer 5

Yikes. I am hopind I didn't do any mistake, but that would be a miracle :)

Let $x= \left( p+q\sqrt{3}\right) ^{1/3}\,;\, y= \left( p-q\sqrt{3}\right) ^{1/3}$.

Then

$$xy= (p^2-3q^2)^\frac{1}{3} $$

Hence

$$2p=x^3+y^3=(x+y)^3-3xy(x+y)=n^3-3n\sqrt[3]{p^2-3q^2}$$

This shows that $\sqrt[3]{p^2-3q^2}$ must be rational, hence integer.

Let $p^2-3q^2=k^3 (*)$. Then $y=\frac{k}{x}$ and thus

$$x+\frac{k}{x} =n \Rightarrow x^2-nx+k=0 \Rightarrow x= \frac{n \pm \sqrt{n^2-4k}}{2} \,.$$

The two roots of this equation must be $x$ and $y$, and thus we get:

$$x=\frac{n + \sqrt{n^2-4k}}{2} \,;\, y= \frac{n - \sqrt{n^2-4k}}{2} \,.$$

Then

$$p+q\sqrt{3}=\left(\frac{n + \sqrt{n^2-4k}}{2} \right)^3 \,.$$

From here, we get that $\sqrt{n^2-4k} \notin \mathbb Q$ and hence

$$p=\frac{n^3+3n^3-12nk}{8}=\frac{n^3-3nk}{2}$$ $$q\sqrt{3}=\frac{3n^2+n^2-4k}{8}\sqrt{n^2-4k}=\frac{n^2-k}{2}\sqrt{n^2-4k}$$

Thus, $n^2-4k=3u^2 $, for some $u$, hence

$$k=\frac{n^2-3u^2}{4} (***)$$

Thus, we get:

$$p=\frac{n^3+9nu^2}{8}$$ $$q=\frac{3n^2+3u^2}{8}u=\frac{3n^2u+3u^3}{8}$$

With this choice we have

$$p+q\sqrt{3}=(\frac{n+u\sqrt{3}}{2})^3$$ $$p-q\sqrt{3}=(\frac{n-u\sqrt{3}}{2})^3$$

Note that $p,q$ integers if and only if $8|n(n^2+u^2)$ and $8|u(n^2+u^2)$. It is easy to check that the first one cannot happen if $n$ is odd. Thus, $n$ must be even and $ 8|u(n^2+u^2) \Rightarrow 2|u^3 \Rightarrow u$ even.

If $n=2s, u=2t$ we get the general solution

$$p=s^3+9st^2$$ $$q=3s^2t+3t^3$$ $$n=2s$$

Note that in this case,

$$\sqrt[3]{p+q\sqrt{3} }=s+t\sqrt{3}$$ $$\sqrt[3]{p-q\sqrt{3} }=s-t\sqrt{3}$$

Answer 6

This kind of simplification occurs if and only if $p \pm q\sqrt d$ has a cube root of the form $x \pm y\sqrt d$ with rational $x,y$. So, to get all instances of this, start by choosing $x+y\sqrt d$, and cube it to get the values for $p$ and $q$.

Setting up the system $(x + y\sqrt d)^3 = p + q\sqrt d$, we get $x^3+3dxy^2 = p$ and $3x^2y+dy^3 = q$. This system has $9$ solutions in $\Bbb C^2$.

To get a particular solution, pick a cube root $a$ of $p + q\sqrt d$, a cube root $b$ of $p - \sqrt d$, and build $x = (a+b)/2, y = (a-b)/2\sqrt d$ : Then, $8(x^3+3dxy^2) = (a+b)^3+3(a+b)(a-b)^2 = 4a^3 + 4b^3 = 8p$, and $8\sqrt d(3x^2y+dy^3) = 3(a+b)^2(a-b)+(a-b)^3 = 4a^3-4b^3 = 8q\sqrt d$, which proves that those $9$ couples are solution.

To show that it has only $9$ solution, we can show $x$ is a root of a degree $9$ polynomial : squaring the second equation we get $q^2 = 9x^4y^2 + 6dx^2y^4 + d^2y^6$. Multiply by $27dx^3$ and use the first equation to get $27dx^3q^2 = 81x^6(p-x^3) + 18x^3(p-x^3)^2 + (p-x^3)^3$, which reduces to $$64x^9-48px^6+(27dq^2-15p^2)x^3-p^3 = 0$$

Finally, given a value for $x$, we can solve for $y$ : The first equation gives $y^2 = \frac{p-x^3}{3dx}$, and plugging this into the second we get $y = \frac{3qx}{8x^3+p}$. Hence the $9$ solutions I gave earlier are the only solutions to the system.

More importantly, if we can find a pair of cube roots such that $x$ is rational, then the corresponding $y$ is also rational, and what we really have found is that $p+q\sqrt d$ has a cube root in $\Bbb Q(\sqrt d)$.

Now, this makes all problems of the form "show that $\sqrt[3]{p+q\sqrt d} + \sqrt[3]{p-q\sqrt d} = 2x$" solvable immediately by computing $y= \frac{3qx}{8x^3+p}$ and then checking that $(x+y\sqrt d)^3 = p + q\sqrt d$. If this doesn't work, then the problem was wrong to begin with.

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I should also point out that the degree $9$ polynomial in $x$ factors over $\Bbb Q(\sqrt{-3},\sqrt[3]{p^2-dq^2})$ into a product of $3$ cubics : $$(4x^3 - 3\sqrt[3]{p^2-dq^2}x - p)(4x^3 - 3j\sqrt[3]{p^2-dq^2}x - p)(4x^3 - 3j^2\sqrt[3]{p^2-dq^2}x - p) = 0$$ (where $j$ is a primitive cube root of $1$).

Attempting to use Cardan's formula on those will give you back the original expression $2x = \sqrt[3]{p+q\sqrt d}+\sqrt[3]{p-q\sqrt d}$.
Attempting to use Cardan's formula on the cubic for $x^3$ and then taking a cube root seems to be an even worse idea, and the moral of the story is : unlike square roots, you can't algebraically find the cube roots of $p + q\sqrt d$.