4
$\begingroup$

Solve in positive integers: $5x^2+6x^3=z^3$.

$x^2(6x+5)=z^3$

  • If $(x,5)=5$, let $x=5k$. So $k^2(6k+1)=\left(\frac{z}{5}\right)^3$, we're left with solving $6n^3+1=m^3$.
  • If $(x,5)=1$, then we're left with solving $6n^3+5=m^3$.

Here we only used $(a,b)=1,\: ab=c^3$ gives $a=n^3, b=m^3$.

How to solve these 'cubic Pell equations'?

$\endgroup$
1
  • 1
    $\begingroup$ AoPS. $\endgroup$
    – user26486
    May 5, 2015 at 17:04

1 Answer 1

3
$\begingroup$

No idea about how to solve this type of 'cubic Pell equations' but the equation in your first case: $$6n^3 + 1 = m^3\tag{*1}$$ doesn't have any non-trivial solution. There is a theorem by Skolem.

For any integer $d \ne 0$, there exists at most one pair $(x,y) \in \mathbb{Z} \times \mathbb{Z}$ with $y \ne 0$ such that $$x^3 + dy^3 = 1\tag{*2}$$

I look at the reference$\color{blue}{^{[1]}}$ which I find this theorem. It says for positive cube free integer $d \le 100$, the only one that $(*2)$ has a (necessarily unique) integer solution with $y \ne 0$ are $$d = 1,2,7,9,17,19,20,26,28,37,43,63,65,91$$

Since $d = 6$ is not on the list, this means neither the equation $x^3 + 6y^3 = 1$ nor $6n^3 + 1 = m^3$ have any non-trivial solution.

For more details, please consult the reference I listed below.

Notes

$\endgroup$
1
  • $\begingroup$ @user236182 $3$ is an odd number and the reference I quote are talking about integral instead of positive integer solutions. One can substitute $(x,y)$ by $(m,-n)$ and apply the result there. $\endgroup$ Sep 10, 2015 at 19:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .