Question: Prove that for any nonnegative integer $n$, $n^3 - n$ is divisible by $3$.
So I suppose that $n^3 - n = 3m$ for some integer $m$.
I know it is true for $n= 1$.
Suppose it is true for $n=k$. So $k^3 - k = 3t$ for some integer $t$.
How to prove it for $n= k+1$? I didn't get the equation for $n= k+1$.
(k+1)^3 - (k+1) = k^3 +3k^2+3k+1 - k - 1
= (k^3 - k) + 3k^2 + 3k
= 3t + 3(k^2 + k)
= 3(t + k^2 + k) which is divisible by 3
since (t+ k^2+k)i s integer
it is hold for n= k+1
Assume it's true that $n^3 - n$ is divisible by $3$. Then
\begin{eqnarray} (n+1)^3 - (n+1) & = & (n+1)\left [(n+1)^2 - 1\right ] \\ & = & (n+1)\left [ n^2 + 2n\right ] \\ & = & n^3 + 3n^2 + 2n \\ & = & \left (n^3 - n\right ) + 3\left (n^2 + n\right ). \end{eqnarray}
Note: $n^3-n=n(n^2-1)=n(n-1)(n+1);$
Base case:
$n=1√$
Hypothesis:
$n(n-1)(n+1)$ is divisible by $3$.
Step for $n+1$:
$(n+1)(n)(n+2) =$
$n(n+1)((n-1)+3)=$
$n(n+1)(n-1) +3n(n+1).$
The first term is divisible by $3$ by hypothesis, the second term is a multiple of $3$, their sum is divisible by $3$.
