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 $x(x+1)\equiv 3\mod 5$?

Note that by Little Fermat,
$$
(x^3)^3\equiv x^{2\phi(5)+1}\equiv x\pmod{5}
$$
Thus, if $x^3\equiv y^3\pmod{5}$, by cubing both sides, we must have $x\equiv y\pmod{5}$. Therefore,
$$
5\mid x^3-y^3\implies5\mid x-y
$$
