Problem 9.4. Solve in integers the equation

$$
x^{3}+10 x-1=y^{3}+6 y^{2}
$$

Sava Grozdev, Svetlozar Doychev
9.4. It is clear that $x$ and $y$ have different parity. Then $k=x-y$ is an odd number and

$$
(3 k-6) y^{2}+\left(3 k^{2}+10\right) y+k^{3}+10 k-1=0
$$

The discriminant of this equation is equal to

$$
D=-3 k^{4}+24 k^{3}-60 k^{2}+252 k+76
$$

and must be a perfect square. Since $D=-k^{2}\left(k^{2}-24 k+60\right)+252 k+76$, then $D<0$ for $k \leq-1$. On the other hand, $D=3 k^{3}(8-k)+2\left(38-k^{2}\right)+$ $2 k(126-29 k)$ and therefore $D<0$ for $k \geq 8$. Since $D=-71<0$ for $k=7$, it remains to check the cases $k=1,3,5$. We have $D=289=17^{2}, D=697$ and $D=961=31^{2}$, respectively, that give the solutions $x=6, y=5$ and $x=2$, $y=-3$.