Austria 2014

We factorise both equations and obtain $$x(x+1)=(y-1)y(y+1),(1)$$ $$y(y+1)=(x-1)x(x+1).(2)$$ We insert (2) into (1) and obtain $$x(x+1)=(y-1)(x-1)x(x+1).(3)$$ 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 $$1=(y-1)(x-1).(4)$$ In particular, we have $x\ne 1$ and $y\ne 1$. This is equivalent to $$y=1+\frac{1}{x-1}=\frac{x}{x-1}⟺y+1=\frac{2x-1}{x-1}.$$ 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 $$\begin{array}{rl}2x-1=(x-1{)}^3(x+1)& ⟺2x-1=(x^2-2x+1)(x^2-1)\\ & ⟺2x-1=x^4-2x^3+x^2-x^2+2x-1\\ & ⟺x^4-2x^3=0⟺x=2.\end{array}$$

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