Mongolian Mathematical Olympiad

Answer: $(a,b)=(0,1)$, $(1,2)$, $(2,2)$ and $(-n,n)$ for any integer $n\ge 0$. If $a+b=0$ then $(a,b)=(-n,n)$ for an integer $n\ge 0$, so now assume that $a+b\ne 0$. Then $a+b=a^2-ab+b^2$ and hence $(a-b{)}^2+(a-1{)}^2+(b-1{)}^2=2$. By setting $x=a-1,y=b-1$ we get $x\le y$ and $x^2+y^2+(x-y{)}^2=2$. Clearly, $(x,y)=(\pm 1,\pm 1)$, $(0,1),(-1,0)$, and therefore $(a,b)=(0,1),(1,2),(2,2)$.