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

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The induction is not as messy as it might look; it's just one application of the binomial theorem (look at difference of $f(x+5)-f(x)$ with $f(x)=x(x+1)$), – Peter Košinár Feb 24 '14 at 2:15
up vote 10 down vote accepted

$$x^2+x\equiv 3\pmod 5\\ \implies x^2+x+4^{-1}\equiv 3+4^{-1}\pmod 5\\ \implies (x+2^{-1})^2=(x+3)^2\equiv 2\pmod 5$$

The quadratic residues modulo $5$ are $0,1,4$ and this list does not include $2$. QED

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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, $$ 5\mid x^3-y^3\implies5\mid x-y $$

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$x^3 - y^3 = (x - y)(x^2 + x y + y^2)$. 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 $x \equiv y \mod 5$.

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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.

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Hint $\ \ x\mapsto x^3\,$ is onto by $\,\{0,1,2,3,4\}^3\!\equiv\{0,1,3,2,4\},\,$ therefore $\,1$-$1$.

Hence $\,x ^3\equiv y^3\,\Rightarrow\, x\equiv y\,\Rightarrow\, 5\mid x-y.\,$ In particular, $\ x-y\ne \pm1.$

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Is surjectivity really easier to see directly than injectivity? – Marc van Leeuwen Feb 24 '14 at 14:05

$x = 5n + k$, 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 $0\pmod 5$.

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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.

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$5$ 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!

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