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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! :)

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  • $\begingroup$ Note $2^3-2=6$, so which gives you $3$ possible options for $m$. $\endgroup$ – Arthur Sep 27 '14 at 19:10
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$n^3=(n-1)n(n+1)$ - 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 $m=6$.

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  • $\begingroup$ Yes, I corrected answer. $\endgroup$ – user165101 Sep 27 '14 at 19:21
  • $\begingroup$ If you're proving this using mathematical induction does that mean that you have to show that $m=6$ is the largest number that divides $k+1$? That is, $m=6$ is the largest number that divides $(k+1)^3 - (k+1)$ ? $\endgroup$ – user45417 Sep 27 '14 at 22:01
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note that $n^3-n=(n-1)n(n+1)$ thus we have $6|(n-1)n(n+1)$ for all $n$ $\in \mathbb{N}$

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  • $\begingroup$ Why is it 6? how do you know just by splitting up n$^3$-n? $\endgroup$ – user174422 Sep 27 '14 at 19:23
  • $\begingroup$ we have three consequent integers, one of them is even and 3 is a divisor of three consequent integers $\endgroup$ – Dr. Sonnhard Graubner Sep 27 '14 at 19:25

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