0

Prove that $$n^3 - n$$ is a multiple of 6 for all positive integral values of n

Does positive integral values of n refer to values of n once the expression is integrated to $$1/4n^4 - 1/2n + c$$ How do you deal with the constant of integration in a proof like this?

FelixM
  • 61

2 Answers2

1

Little fermat implies that $n^3=n$ mod $3$ and $n^3-n=n(n^2-1)=n(n-1)(n+1)$ and $n(n-1)$ is even. $n-1,n,n+1$ are 3 consecutive numbers, one of them is divisible by $3$.

Postive integral value are elements of $\mathbb{N}$ strictly positive.

Tsemo Aristide
  • 87,475
  • 1
    There is no need to invoke Fermat, one even and one multiple of $3$ are enough. –  Jul 06 '20 at 09:54
1

$$n^3-n\equiv((n\bmod3)^3-(n\bmod3))\mod6$$

and

$$0^3-0=0,1^1-1=0,2^3-2=6.$$