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

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

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

\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).

Attribution
Source : Link , Question Author : Américo Tavares , Answer Author : Ian Mateus

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