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

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


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

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?

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    $\begingroup$ Noting that $26/15$ is a convergent to $\sqrt 3$ I found $(7+5\sqrt 2)^{(1/3)}+(7-5\sqrt 2)^{(1/3)}=2$. Emboldened, I tried the next two convergents, $(17,12)$ and $(41,29)$. Failure. $\endgroup$ – Ross Millikan Apr 27 '13 at 21:04
  • $\begingroup$ @Ross Millikan Your solutions are for a different problem. I restricted it to numbers of the form $p+q\sqrt{3}$, not $p+q\sqrt{r}$. $\endgroup$ – Américo Tavares Apr 27 '13 at 21:28
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    $\begingroup$ I find $(1351+780\sqrt 3)^{(1/3}+(1351-780\sqrt 3)^{1/3}=14$, also $(70226,40545,52), (3650401,2107560,194), (189750626,109552575,724)$ Every third covergent seems to work. $\endgroup$ – Ross Millikan Apr 27 '13 at 21:40
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    $\begingroup$ @CameronBuie: they are the fractions that come from the continued fraction for $\sqrt 3$. They are also solutions to Pell's equation $x^2-3y^2=1$ Given one pair $(a,b)$, the next one is $(2a+3b,a+2b)$, so they start $(1,0), (2,1), (7,4), (26,15)$ and the ratio approximates $\sqrt 3$ quite well. $\endgroup$ – Ross Millikan Apr 27 '13 at 21:53
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    $\begingroup$ IIRC, the family of problems "does $a + b\sqrt d$ have a cube root in $\Bbb Q(\sqrt d)$" and "does $x^3 + px + q"$ have a rational root ?" are exactly as hard : given one instance of one kind of problem there is an instance of the second kind such that they either both have solutions or both don't, and in case they do, there is a formula to get the answer to the first problem from the answer to the second problem. And that goes both way. $\endgroup$ – mercio Apr 28 '13 at 12:31

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.

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 $$\left(\frac{2589437}{8}+\frac{56351}{4}\sqrt{11}\right)^{1/3}+\left(\frac{2589437}{8}-\frac{56351}{4}\sqrt{11}\right)^{1/3}=137$$


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

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    $\begingroup$ This is both simpler and more rigorous than my solution, and seems to find exactly the same family of solutions. (+1) $\endgroup$ – Micah Apr 27 '13 at 23:30
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    $\begingroup$ I still have to analyze your last edit. Overall it seems impressive. $\endgroup$ – Américo Tavares Apr 28 '13 at 18:47

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 $$ \left(\frac{1885}{4}+\frac{1431}{4} \sqrt{3}\right)^{1/3}+\left(\frac{1885}{4}-\frac{1431}{4} \sqrt{3}\right)^{1/3}=5 \, . $$

  • $\begingroup$ +1 Thanks! It seems to me that there are only two open points: 1. How to generate integer solutions? 2. The proof mentioned in the first paragraph. $\endgroup$ – Américo Tavares Apr 27 '13 at 22:57
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    $\begingroup$ @AméricoTavares: I think Ian's answer gives the proof I want (his $8p/n-n^2$ is my $m^2$). As for integer solutions, one start would be to notice that if $p,q,n$ are to be integers, $m$ must be an integer multiple of $\frac{1}{n}$. For any fixed $n$, finding which multiples of $\frac{1}{n}$ lead to integer $p,q$ is a matter of modular arithmetic, so you can do it by process of elimination if nothing else. I'm not sure if there's an approach that'd work for general $n$ or not... $\endgroup$ – Micah Apr 27 '13 at 23:27
  • $\begingroup$ Wouldn't it be easier just to pick naturals $a,b$ and form $p+q\sqrt 3=(a+b \sqrt 3)^3?$ This gives $p=a^3+9ab, q=b^3+3ab$ $\endgroup$ – Ross Millikan Apr 28 '13 at 1:15

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.

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 $p + q\sqrt d$.

  • $\begingroup$ Thanks for developing the ideas of your comment above! In the last equation is $j$ a constant to be found by equating their coefficients to the coefficients of the degree 9 equation in $x$ (or another adequate technique)? $\endgroup$ – Américo Tavares Apr 30 '13 at 9:19
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    $\begingroup$ @AméricoTavares : $j$ is a primitive cube root of $1$. $\endgroup$ – mercio Apr 30 '13 at 9:20
  • $\begingroup$ Ah! Thanks, I've not seen that. $\endgroup$ – Américo Tavares Apr 30 '13 at 9:21
  • $\begingroup$ You claim "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 $x$, $y$." The if part is easy, but how should we proceed about the only if case? This seems non trivial. $\endgroup$ – Ian Mateus Feb 1 '14 at 17:52
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    $\begingroup$ @IanMateus : in the answer I show that the $9$ complex solutions to $(x+y\sqrt d)^3 = (p+q\sqrt d)$ satisfy $y = 3qx/(8x^3+p)$ and $x$ is the arithmetic mean of a cube root of $p+\sqrt d$ and a cube root of $p-\sqrt d$. In particular if you find two roots that sum to a rational, then the corresponding $x$ and $y$ are both rational, so that $p+q \sqrt d$ had a cube root in $\Bbb Q(\sqrt d)$ all along. If you're still not happy, you can check that in $K(p,q,d,r,a,b,x)/(r^2-d,a^3-p-qr,b^3-p+qr,2x-a-b), (x+\frac{3qx}{8x^3+p}r)^3 = p+qr$. $\endgroup$ – mercio Feb 2 '14 at 13:37

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.

  • $\begingroup$ So my equation $(2)$ has infinitely many solutions! $\endgroup$ – Américo Tavares Apr 27 '13 at 22:15
  • $\begingroup$ @AméricoTavares: Yes it does. I don't know where to look for more. $\endgroup$ – Ross Millikan Apr 27 '13 at 22:17
  • $\begingroup$ Thanks! It may well be there is no additional family of solutions. $\endgroup$ – Américo Tavares Apr 27 '13 at 22:20
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    $\begingroup$ @AméricoTavares: Micah has done much better. He has a doubly infinite family. I have confirmed it includes mine. $\endgroup$ – Ross Millikan Apr 27 '13 at 22:27

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

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


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



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:

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


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

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,

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


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.


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