, 2014

We factorise both equations and obtain
$$x(x + 1) = (y - 1)y(y + 1), \quad (1)$$
$$y(y + 1) = (x - 1)x(x + 1). \quad (2)$$
We insert (2) into (1) and obtain
$$x(x + 1) = (y - 1)(x - 1)x(x + 1). \quad (3)$$
We first consider the cases $x = 0$ and $x = -1$, where (2) results in $y \in \{0, -1\}$. Thus we obtained the solutions
$$(0, 0), (0, -1), (-1, 0), (-1, -1).$$

From now on, we assume that $x \notin \{0, -1\}$. Cancelling $x(x + 1)$ in (3) yields
$$1 = (y - 1)(x - 1). \quad (4)$$
In particular, we have $x \neq 1$ and $y \neq 1$. This is equivalent to
$$y = 1 + \frac{1}{x-1} = \frac{x}{x-1} \iff y + 1 = \frac{2x-1}{x-1}.$$
We insert this into (2) and obtain
$$\frac{x}{x-1} \cdot \frac{2x-1}{x-1} = x(x-1)(x+1)$$
which is equivalent to
$$\begin{align*} 2x - 1 = (x - 1)^3 (x + 1) &\iff 2x - 1 = (x^2 - 2x + 1)(x^2 - 1) \\ &\iff 2x - 1 = x^4 - 2x^3 + x^2 - x^2 + 2x - 1 \\ &\iff x^4 - 2x^3 = 0 \iff x = 2. \end{align*}$$

because $x \neq 0$. Inserting this in (4) yields $y = 2$, so we got the fifth solution $(2, 2)$. Therefore, all solutions are given as
$$(0, 0), (0, -1), (-1, 0), (-1, -1), (2, 2).$$