1

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.

Jordan Glen
  • 1,711
Ldog327
  • 519

3 Answers3

6

$$a^3-a=(a-1)a(a+1)$$ 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)

Community
  • 1
lab bhattacharjee
  • 274,582
1

Consider the following options:

  • $a\equiv0\pmod6 \implies a^3-a\equiv 0-0\equiv6\cdot 0\equiv0\pmod6$
  • $a\equiv1\pmod6 \implies a^3-a\equiv 1-1\equiv6\cdot 0\equiv0\pmod6$
  • $a\equiv2\pmod6 \implies a^3-a\equiv 8-2\equiv6\cdot 1\equiv0\pmod6$
  • $a\equiv3\pmod6 \implies a^3-a\equiv 27-3\equiv6\cdot 4\equiv0\pmod6$
  • $a\equiv4\pmod6 \implies a^3-a\equiv 64-4\equiv6\cdot10\equiv0\pmod6$
  • $a\equiv5\pmod6 \implies a^3-a\equiv125-5\equiv6\cdot20\equiv0\pmod6$
barak manos
  • 43,109
0

by Euler's theorem $ a^{\phi (n)} = a (mod n) $ with $\phi(n)$ the number of integers lower than n and prime with n if n = 6 there are three integers less than 6 and first ones with 6: 1,3,5 let $a^3 = a (mod 6) $

curieux_2014
  • 316
  • 1
  • 6