Prove that for every integer $n$, $n^3$mod$6$=$n$mod$6$

Question

I will use induction to prove this. Firstly for $n=1$,
$1^3\text{mod}6=1\text{mod}6$

Now we assume that this holds for some $n=k$ and prove that if it holds for $n=k$ it should also hold for $n=k+1$.

$(k+1)^3\text{mod}6=(k^3+3k^2+3k+1)\text{mod}6$

As for $n=k$ because it holds,

$k^3\text{mod}6=k\text{mod}6$

$k^3-6p(k)=k-6q(k)$

$k^3-k=6p(k)-6q(k)$

Consider $(k^3+3k^2+3k+1)-6a(k)=6p(k)-6q(k)+k+1+3k^2+3k$

Now we just need to prove that $3k^2+3k$ is always a multiple of 6.

For $k=1$ it holds.

Assume $k=j$ holds. Then for $k=j+1$, $3(j+1)^2+3(j+1)=3(j^2+2j+1)+3j+3=3j^2+3j+6j+6$
So for $k=j+1$ it is also a multiple. As it holds for $k=1$, then it holds for all positive integers $k$.

Then this means that
$6p(k)-6q(k)+k+1+3k^2+3k=k+1-6b(k)$

And if $(k^3+3k^2+3k+1)-6a(k)=k+1-6b(k)$ this means $(k+1)^3\text{mod}6=(k+1)\text{mod}6$

Is this correct?

Answer 1

An easy approach:

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

As in any two consecutive integers, one is even, and in any three consecutive, one is multiple of three. So the product of three consecutive numbers is a multiple of $6$. $$\begin{split}n^3-n&\equiv0\pmod6\\n^3&\equiv n\pmod 6\end{split}$$ Done!

Answer 2

Why are you inducting in this case? I mean if we are working in $\mathbb Z/6\mathbb Z$, then there is no need to check it for all integers $n$. Anyways, this is just a special case of Euler's Theorem.