Solve the equation in integers: $x^{3}+3=4 y(y+1)$.
Let's write the equation as $x^{3}=(2 y+1)^{2}-4=(2 y-1)(2 y+3)$. The odd numbers $2 y-1$ and $2 y+3$ are coprime, since their difference is 4. Therefore, both of them must be perfect cubes. However, the difference between two cubes cannot be 4.

## Answer

No solutions.

Send a comment