Solve the given equation in integers \begin{align*} y^3=8x^6+2x^3y-y^2 \end{align*}
Problem
Source: Balkan MO ShortList 2009 N1
Tags: number theory
07.04.2020 05:14
Consider the equation a quadratic in $x^3$ and we will get : $x^3=\frac{n^3+n^2-2n}{8}$ and $x^3=\frac{-(n + 1)(n - 1)(n + 2)}{8}$ $y=\frac{n(n+1)}{2}-1$ For some $n\in \mathbb{Z}$
10.04.2020 07:08
The only solutions of the Diophantine equation $(1) \;\; y^3 = 8x^6 + 2x^3y - y^2$ are $(x.y)=(0,-1), (0,0), (1,2)$. Proof: Equation (1) give us $(2x)^3 = y(-1 \pm \sqrt{8y + 9})$. Hence by equation (2) there is a non-negative integer $n$ s.t. $8y + 9 = (2n + 1)^2$, i.e. $(3) \;\; {\textstyle y = \frac{n(n + 1)}{2} - 1}$, which inserted in equation (3) result in $(4) \;\; {\textstyle (2x)^3 = (\frac{n(n + 1)}{2} - 1)(-1 \pm (2n + 1))}$. Consequently by equation (4) we obtain $(5) \;\; (2x)^3 = n(n - 1)(n + 2)$ or $(6) \;\; (2x)^3 = -(n + 1)(n - 1)(n + 2)$. Set $f(n) = n(n - 1)(n + 2)$ and $g(n) = -(n + 1)(n - 1)(n + 2)$. Then $f(n) = g(-n-1)$, which means in order to solve equations (5) and (6), we have to find those $n$ for which $f(n)$ is a perfect even cube. Hence by equation (6) we have $(7) \;\; n^3 + n^2 - 2n = (2x)^3$, which combined with the fact that $n^3 < n^3 + n^2 - 2n < (n + 1)^3$ for all $|n| \geq 3$. Therefore the equation (7), i.e. $(2x)^3 = f(n)$ is solvable only if $|n| \leq 2$. Seeing that $f(n)=0$ when $n \in \{-2,0,1\}$, $f(-1)=2$ and $f(2)=8$, we deduce $f(n)=(2x)^3$ iff $x=0$ or $x=1$. Therefore the solutions of equation (5) are $(x,n) = (0,-2), (0,0), (0,1),(1,2)$ which means (since $f(n)=g(-n-1)$) the solution of equation (6) are $(x,n) = (0,-2), (0,-1),(0,1),(1,-3)$. These six different pairs $(x,n)$ combined with the formula (3) give us the following three solutions of equation (1): $(x,y) = (-1,0), (0,0), (1,2)$. q.e.d.