Variation of Mordell's equation Determine all integral solutions to the equation $$3x^2+4=y^3$$ I'm not even sure where to start with this.

Hint : Considering in mod $3$ helps.

> We have to have $y\equiv 1\pmod 3$. Let $y=3k+1$ where $k\in\mathbb Z$. Then, the equation can be written as $x^2=3(3k^3+3k^2+k)-1$. So $x^2\equiv 2\pmod 3$, which is impossible.