Proving divisibility of $a^3 - a$ by $6$ [duplicate]

divisibility

As part of a larger proof, I need to show why $a^3-a$ is always divisible by $6$. I'm having trouble getting started.


$$a^3-a=(a-1)a(a+1)$$ which is a product of three consecutive integers.

So, exactly one of them must be divisible by $3$

and at least one of them must be even.

More generally The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)


Consider the following options:

  • $a\equiv0\pmod6 \implies a^3-a\equiv 0-0\equiv6\cdot 0\equiv0\pmod6$
  • $a\equiv1\pmod6 \implies a^3-a\equiv 1-1\equiv6\cdot 0\equiv0\pmod6$
  • $a\equiv2\pmod6 \implies a^3-a\equiv 8-2\equiv6\cdot 1\equiv0\pmod6$
  • $a\equiv3\pmod6 \implies a^3-a\equiv 27-3\equiv6\cdot 4\equiv0\pmod6$
  • $a\equiv4\pmod6 \implies a^3-a\equiv 64-4\equiv6\cdot10\equiv0\pmod6$
  • $a\equiv5\pmod6 \implies a^3-a\equiv125-5\equiv6\cdot20\equiv0\pmod6$

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