Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these statements imply that $x(x+1)\equiv 3\mod 5$. Thus I can finish this by showing that there are no such integers which satisfy $x(x+1)\equiv 3\mod 5$.
I want to say that it is sufficient to check by hand for the values 0,1,2,3, and 4 (for which it is not true), but other than following this by a messy induction I was wondering if there is an easier way to show that there are no integers such that $x(x+1)\equiv 3\mod 5$?

$x^3 - y^3 = (x - y)(x^2 + x y + y^2)$.  Now 
$x^2 + x y + y^2 \equiv (x + 3 y)^2 - 3 y^2 \mod 5$, and since $3$ is not a quadratic residue mod $5$ we find that there are no solutions to 
$x^2 + x y + y^2 \equiv 0 \mod 5$ other than the trivial $(0,0)$.  Thus
$x^3 \equiv y^3 \mod 5$ only when $x \equiv y \mod 5$.
