229. Cubic Diophantine Equation. Solve the equation in integers

$$
x^{3}+1=y^{2}
$$
229. One solution is obvious: $(x, y)=(-1,0)$. Moreover, it is easy to notice that $x$ cannot take other negative values, since $y^{2}$ is non-negative. Let's write our equation as

$$
\begin{aligned}
& x^{3}+1=(x+1)\left(x^{2}-x+1\right)= \\
& =(x+1)^{2}\left(x-2+\frac{3}{x+1}\right)=y^{2}
\end{aligned}
$$

Further, the expression $\frac{3}{x+1}$ will be an integer only when $x=0$ or $x=2$. From this, we can find the remaining four solutions: $(0, \pm 1)$ and $(2, \pm 3)$. There are no other solutions.