There are no more than $p$ elements divisible by $p$

Question

Let $p$ is prime number has the form $3k+2$. Prove that in the set $$S=\{ y^6-x^3-2\mid x,y\in \mathbb{Z}:0\le x\le p-1, 0\le y\le p-1 \}$$ there are no more than $p$ elements divisible by $p$.

I really have no idea to solve it. Help me.  

Answer 1

Since $p-1$ is not divisible by $3$, the field $\mathbb F_p$ does not contain any non-trivial roots of unity. As a result, the mapping $z\mapsto z^3$ is a bijection of $\mathbb F_p$; that is, any element of $\mathbb F_p$ can be uniquely represented as a cube. Consequently, any element of $\mathbb F_p$ has a unique cube root.

To show that your set $S$ contains at most $p$ elements divisible by $p$, just notice that there are at most $p$ pairs $(x,y)\in\mathbb F_p\times\mathbb F_p$ with $y^6-x^3-2=0$: namely, for every given $y\in\mathbb F_p$, we are forced to take $x$ to be the cube root of $y^6-2$.