Find the largest natural number m such that n$^3$-n is divisible by m for all n$\in$ $\mathbb{N}$. Prove your assertion.
So my basis that I have is: Notice that (1)$^3$-(1)=0, and m(0)=0, so m divides into n$^3$-n.
I tried to start the inductive step, but i'm getting a little lost, because i'm so used to doing this problem where m is a definitive number. So I have,
Fix some integer k greater or equal to 1, and suppose k$^3$-k is divisible by m. Then, there exists an integer s for which k$^3$-k=ms.
I'm not really sure if all of that was correct, but if it was the way I was supposed to start this, then what do i do next? would I then plug in k+1 with k? And if so, then I still don't know how to find what m is.
Please help! 🙂
Best Answer
$n^3=(n-1)n(n+1)$ - product of three consequent integers. At least one of them is even, so $2|(n^3-n)$. They all have different residues modulo $3$, so one of them is divisible by $3$, so $3|(n^3-n)$. So, $6|(n^3-n)$ for every $n$, so $m\geq 6$. Let's prove that $m=6$. Let $k$ be a natural number, such that $k|(n^3-n)$ for every $n$. If $k$ has a prime divisor $p\neq 2,3$ then $k\not|(p+2)^3-(p+2)=(p+1)(p+2)(p+3)$ since $p<p+1,p+2,p+3<2p$. So, $k$ is of the form $2^a3^b$. $a$ can't be greater than $1$ because $4\not|2^3-2$ and $b$ can't be greater than $1$ because $9\not|2^3-2$. So, $k\leq2^13^1=6$ and $m=6$.