Show if $n \equiv 3$ (mod 7) then $x^3 +y^3 =n$ has no solution.

Question

Show if $n \equiv 3$ (mod 7) then $x^3 +y^3 =n$ has no solution.

I know that $n \equiv 3$ (mod 7) implies $7 | n-3$ which implies $n = 7m+3, m\in\mathbb{Z}$.

So $x^3 + y^3 = 7m+3$ which means $ x^3 +y^3 -3 =7m$. I am not sure where to go from here.

Answer 1

The cubes in $\mathbf Z/7\mathbf Z$ are $0$, $1$ and $-1$. Hence the sums of two cubes are $\;\bigl\{0,\, \pm 1,\,\pm2\bigr\}$. Neither $3$ nor $-3\,$ are in this list.