, 2011

If $y = 0$ or $y = -1$, the right hand side of the given equation is equal to zero so $x^3 + x^2 + x = x(x^2 + x + 1) = 0$, and hence $x = 0$. That gives us two solutions: $(0, 0)$ and $(0, -1)$.

We shall prove that no other solutions exist.

Assume that $y \in \mathbb{Z} \setminus \{-1, 0\}$. Then $y^2 + y > 0$ so $x(x^2 + x + 1) > 0$, and it follows that $x > 0$.

If $(x, y)$ is the solution of the given equation, then $(x, -y-1)$ is also the solution since $y^2 + y = (-y - 1)^2 + (-y - 1)$. In other words, if there are no solutions for $y > 0$, then there are no other solutions. Therefore, we can assume that $y > 0$.

The given equation can be written as:
$$x^3 = (y - x)(x + y + 1).$$
Let us prove that $\gcd(y - x, x + y + 1) = 1$. Assume the contrary, let $p$ be the common prime divisor of the numbers $y - x$ and $x + y + 1$.

From the above we conclude that $p \mid x^3$ and hence $p \mid x$. Now we have:
$$p \mid y - x, \quad p \mid x + y + 1, \quad p \mid x \quad \implies \quad p \mid (x + y + 1) - (y - x) - 2x = 1,$$
which means that our assumption was wrong, i.e. $y - x$ and $x + y + 1$ are relatively prime numbers whose product is $x^3$ so each of them is a cube of an integer. Furthermore, $x^3 > 0$ and $x + y + 1 > 0$ so $y - x > 0$, which means that $y - x$ and $x + y + 1$ are cubes of positive integers. Let us denote:
$$y - x = a^3 \\ x + y + 1 = b^3.$$
Notice that from $2x + 1 > 0$ it follows that $y - x < x + y + 1$, that is $a < b$, i.e. $b - a \ge 1$. Subtracting the previous two equations we get $2x + 1 = b^3 - a^3$, and from above it follows $x = ab$. Therefore, $2ab + 1 = b^3 - a^3$. Now:
$$2ab + 1 = b^3 - a^3 = (b - a)(a^2 + ab + b^2) \ge a^2 + ab + b^2 \ge 3ab,$$
i.e. $ab \le 1$, which is clearly not possible since $a$ and $b$ are distinct positive integers. Hence, the only solutions of the given equation are $(0, 0)$ and $(0, -1)$.