I have the following statement.
Find the largest natural number $m$ such that $n^3 - n$ is divisible by $m$ for all $n$ in $\mathbb{N}$. Prove your result.
I have deduced that the greatest natural number $m$ would be $6$ and I am pretty sure that I have to prove this statement through induction. T he base case is pretty obvious but it is the inductive step that is giving me trouble.
Please help!
$$n^3 - n = (n-1) \cdot n \cdot (n+1) $$
So, if $m$ has a divisor greater than 3, then for $n=m+2$, $m$ does not divide $n^3-n$.
You only need to eliminate powers of $2$ and $3$
From $2^3 - 2 = 6$ it indeed clearly follows that $m$ can be at most $6$. A prove with induction that $6$ divides $n^3-n$ could go as follows:
Base case: For $n = 1$ it holds, because $6$ divides $1^3-1 = 0$.
Inductive step: Suppose it holds for $n = k$ for some $k \in \mathbb{N}$, so $6$ divides $k^3-k$. Now we have that $$(k+1)^3-(k+1) = k^3 + 3k^2 + 2k = (k^3-k) + 3k(k+1)$$ By the induction hypothesis, we have that $6$ divides $k^3-k$. We also have that $k(k+1)$ is even, because either $k$ or $k+1$ is, so $3k(k+1)$ is also divisible by 6. This means that also $(k^3-k) + 3k(k+1) = (k+1)^3-(k+1)$ is divisible by 6, proving the statement for $n = k+1$.
A faster prove could however use the following factorization: $n^3-n = (n-1)\cdot n \cdot (n+1)$. Now the right-hand side is the product of three consecutive numbers, one of which is divisible by 3, and also at least one of these numbers is divisible by 2, making the expression divisible by 6.
When $n=0$ or $n=1$ we have $n^3-n=0$, which is divisible by any natural number. So consider $n=2$, then $n^3-n=6$, which is divisible by $1,2,3,6$. We will show that $m=6$ is the desired number. Now we use the induction hypothesis that $6\mid n^3-n$ and we want to show that $6\mid (n+1)^3-(n+1)$. Thus, the induction hypothesis tells use that $2\mid n^3-n$ and $3\mid n^3-n$. We have $$\begin{align}(n+1)^3-(n+1)&=(n+1)((n+1)-1)((n+1)+1)\\&=n(n+1)(n+2)\\&\equiv n(n+1)(n-1)\pmod 3\\&\equiv n^3-n\pmod 3\\&\equiv 0\pmod 3\qquad\qquad\qquad\text{since $3\mid n^3-n$}\end{align}$$ and $$\begin{align}(n+1)^3-(n+1)&=(n+1)((n+1)-1)((n+1)+1)\\&=n(n+1)(n+2)\\&\equiv n(n-1)(n)\pmod 2\\&\equiv n^3-n^2\pmod 2\\&\equiv n^3-n\pmod 2\qquad\qquad\qquad\text{since $n^2\equiv n\pmod 2$}\\&\equiv 0\pmod 2\qquad\qquad\qquad\qquad~\text{since $2\mid n^3-n$}\end{align}$$ Hence, $2\mid (n+1)^3-(n+1)$ and $3\mid (n+1)^3-(n+1)$, so we're done, because if $a\mid x$ and $b\mid x$ where $\gcd(a,b)=1$ we have $ab\mid x$.
