Cubic diophantine equation in two variables

Question

I am trying to solve the Diophantine equation $xy^2 + 2xy + x - 243y = 0$.

I simplified it to $x(y^2 +2y +1) = 243y$ but I am stuck on what to do now. Any help would be appreciated.

Answer 1

Do you have to find solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$? The pair $(0,0)$ is a solution.

You can simplify further $x(y+1)^2=243y$.

We know that $y\mid x(y+1)^2$. But $\operatorname{gcd}(y,(y+1)^2)=1$. You can proof that simply.

So $y\mid x$. And we can express $x=yc$. Which would lead to a cubic equation in $y$, which you can solve, as $y$ now accours in every term, we even can factor this, which then leads to a quadratic equation. This can be solved easily with the known formulas/methods.

Answer 2

If $x$ and $y$ are integers satisfying your equation, then also $$3^5y=243y=x(y^2+2y+1)=x(y+1)^2.$$ Because $y$ and $y+1$ have no common divisor, it follows that $(y+1)^2$ divides $3^5$. Can you finish from here?

It follows that $(y+1)^2=3^{2k}$ for some $k\in\{0,1,2\}$ and so $y=-1\pm3^k$, and $$x=\frac{3^5y}{(y+1)^2}=3^{5-2k}y.$$ Checking these few values of $k$ yields the following six solutions $(x,y)$: $$(0,0),\quad(-486,-2),\quad(54,2),\quad(-108,-4),\quad(24,8),\quad(-30,-10).$$

Answer 3

$$xy^2 + 2xy + x - 243y =(x)y^2+(2x-343)y+x=0 \implies x=0,y=0$$ $$y = \frac{(243-2x) \pm 9 \sqrt{3} \sqrt{243 - 4 x}}{(2 x)} \quad\text{for}\quad -1500\le x \le 1500\\ \implies (x,y)=(-486,-2)\quad (-108,-4)\quad (-30,-10)\quad (24,8)\quad (54,2)\quad $$