Iranian Mathematical Olympiad

We have $$\begin{array}{rl}\frac{x^3-xy+1}{x^2+x-y}& =\frac{y^3+xy-1}{y^2+x-y}\Rightarrow x-\frac{x^2-1}{x^2+x-y}=y+\frac{y^2-1}{y^2+x-y}\\ \Rightarrow x-y& =\frac{x^2-1}{x^2+x-y}+\frac{y^2-1}{y^2+x-y}.(\ast )\end{array}$$ Now we have three cases: Case 1. $$x=y\Rightarrow 2\frac{x^2-1}{x}=0\Rightarrow x^2=1\text{ or }x=0\Rightarrow (x,y)=(0,0),(1,1),(-1,-1)$$

Case 2. $x>y$. Let $k=x-y>0$ so by () we have $k=\frac{x^2-1}{x^2+k}+\frac{y^2-1}{y^2+k}$. If $|x|\le 1$ or $|y|\le 1$ by using main equation we get $(x,y)=(-1,0),(1,2),(0,1),(-2,-1)$. Now we can assume $|x|,|y|>1$ $$x^2-1<x^2+k\Rightarrow 0<\frac{x^2-1}{x^2+k}<1\text{ and similarly }0<\frac{y^2-1}{y^2+k}<1,\text{ therefore}$$ $$0<k=\frac{x^2-1}{x^2+k}+\frac{y^2-1}{y^2+k}<2\Rightarrow k=1\Rightarrow 1=\frac{x^2-1}{x^2+1}+\frac{y^2-1}{y^2+1}.(\ast \ast )$$

Case 3. $x<y$. Let $t=y-x>0$ so by () we have $t=\frac{x^2-1}{t-x^2}+\frac{y^2-1}{t-y^2}$ and by adding $2=\frac{t-x^2}{t-x^2}+\frac{t-y^2}{t-y^2}$ to both sides we get $t+2=\frac{t-1}{t-x^2}+\frac{t-1}{t-y^2}$. We claim that $t-x^2$ and $t-y^2$ are not positive, simultaneously. Assume by contrary $t-x^2>0,t-y^2>0$ then $$t>x^2,t>y^2\Rightarrow 2(y-x)=2t>x^2+y^2\Rightarrow (x+1{)}^2+(y-1{)}^2<2.$$ This contradicts because $|x|,|y|>1$. Therefore at least one of fractions $\frac{t-1}{t-x^2}$ and $\frac{t-1}{t-y^2}$ is less than or equal to 0 and the other less than or equal to $t-1$. So $$t+2=\frac{t-1}{t-x^2}+\frac{t-1}{t-y^2}\le 0+(t-1)<t.$$ This contradiction shows that case 3 does not have a new solution and these are all the solutions: $(0,0),(1,1),(-1,-1),(-1,0),(1,2),(0,1),(-2,-1)$. □