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.