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 $x(x+1)\equiv 3\mod 5$?
 A: 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,
$$
5\mid x^3-y^3\implies5\mid x-y
$$
A: $x^3 - y^3 = (x - y)(x^2 + x y + y^2)$.  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 $x \equiv y \mod 5$.
A: Since you are considering things like $(x+1)^3-x^3$, I assume the problem should be to show that the difference of two consecutive cubes is never divisible by $5$.
You could finish the problem in this way:
$$\eqalign{
  x(x+1)\equiv3\pmod5\quad
  &\Rightarrow\quad 4x(x+1)+1\equiv 3\pmod5\cr
  &\Rightarrow\quad (2x+1)^2\equiv 3\pmod5\cr}$$
which implies that $3$ is a square (the technical term is "quadratic residue") modulo $5$.  But it isn't, so this is impossible.
How do you know that $3$ is not a square modulo $5$?
Method $1$: calculate $0^2,1^2,\ldots,4^2$ modulo $5$ and observe that you never get $3$.  This is no easier than what you suggested!
Method $2$: use the Legendre symbol and quadratic reciprocity.  This is probably more work than what you suggested, though if you had a larger modulus instead of $5$ it would probably be better.
A: $x = 5n + k$, then $x^2 + x - 3 = (5n + k)^2 + 5n + k - 3 = k^2 + k - 3\pmod 5$ now check $k = 0, 1, 2, 3$, and $4$ you don't get $0\pmod 5$. 
A: Hint $ \ x\mapsto x^3\,$ is onto by $\,\{0,\pm1,\color{#c00}{\pm2}\}^{\large \color{#c00}3}\!\equiv\{0,\pm1,\color{#c00}{\mp2}\}\pmod{\!5}\,$ so $\,1$-$1\ $ (pigeonhole)
Hence  $\,x ^3\equiv y^3\,\Rightarrow\, x\equiv y\,\Rightarrow\, 5\mid x-y.\,$ In particular, $\ x-y\ne \pm1.$
A: If you've already checked 0,1,2,3,4 by hand, you don't have to do any induction, because $x(x+1) \equiv y(y+1)$ for some $y \in [0..4]$. Likewise, couldn't you do exactly the same thing on the original question? Cubes mod 5 are just as easy to compute as squares mod 5.
A: $$x^2+x\equiv 3\pmod 5\\
\implies x^2+x+4^{-1}\equiv 3+4^{-1}\pmod 5\\
\implies (x+2^{-1})^2=(x+3)^2\equiv 2\pmod 5$$
The quadratic residues modulo $5$ are $0,1,4$ and this list does not include $2$.  QED
A: $5$ is a really small number: checking all the values mod $5$ is the easy way.
You could have even avoided doing all of the work you had done, by just listing all possible cubes mod $5$, and seeing that no consecutive cubes are equal.


*

*$0^3 = 0$

*$1^3 = 1$

*$2^3 = 8 = 3$

*$3^3 = 9 \cdot 3 = 4 \cdot 3 = 2$

*

*or $3^3 = (-2)^3 = -2^3 = -3 = 2$


*$4^3 = \cdots = 4$

*$0^3 = 0$



Notice that every number was a cube; if we had already known that, it would have been obvious no two consecutive cubes could be equal mod $5$. We could prove a theorem:

(Assuming $m > 1$) if every number modulo $m$ is an $n$-th power, then no integer $n$-th powers can be divisible by $m$

One simple sufficient condition for this is if $\gcd(\varphi(m), n) = 1$. This would let us quickly prove similar facts for very large moduli and exponents!
