Hrvatska 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: $$p\mid y-x,p\mid x+y+1,p\mid x⟹p\mid (x+y+1)-(y-x)-2x=1,$$ 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: $$\begin{gathered} y-x=a^3 \\ x+y+1=b^3. \end{gathered}$$ 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 $(0,-1)$.