Selected Problems from the Final Round of National Olympiad

First assume $x>y$. Then $3xy+1=x^3-y^3=(x-y)(x^2+xy+y^2)\ge (x-y)\cdot 3xy$. Thus $3xy=x^3-y^3-1\ge 1-1=0$ because $x>y$. If $3xy>0$, then $3xy\ge 3$, hence the inequality $3xy+1\ge (x-y)\cdot 3xy$ derived above implies $x-y=1$. If $3xy=0$, then either $x=0$ or $y=0$ and in both cases the only possibility is $x-y=1$ again. An elementary check shows that all pairs $(x,y)=(n+1,n)$, where $n$ is an integer, satisfy the initial equation.

Now assume $x=y$. Then the equation has no solutions, since the l.h.s. is 0 while the r.h.s. is positive.

Finally assume $x<y$. Then the l.h.s. of the equation is negative, showing that $xy$ is negative. Hence $x<0$ and $y>0$. Denoting $-x=z$ and multiplying the equation by $(-1)$ leads to new equation $z^3+y^3=3zy-1$. Then $3zy-1=z^3+y^3=(z+y)(z^2-zy+y^2)\ge (z+y)zy$. Hence $z+y<3$, giving $z=y=1$, i.e., $x=-1,y=1$, as the only possibility. It is easy to check that this satisfies the equation.