# If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$

I am trying to teach myself mathematics (I have no access to a teacher), but I am not getting very far. I am just working through the exercises at the end of the book's chapter, but unfortunately there are no solutions.

Anyway, I am trying to prove

If $n = m^3 - m$ for some integer $m$, then $n$ is a multiple of $6$.

But I do not know how to approach it. I thought of starting with something like $n = 6k$ for the multiple and that $m^3$ is crucial, but I do not know how that would help or where to go next. Does anyone have any hints or suggestions? Please do not post the whole proof because I want to solve it myself, thank you.

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You might also want to show that if $n = m^5-m$ for some integer $m$, then $n$ is a multiple of 30. (This is a bit trickier but basically the same idea.) – Michael Lugo Jul 17 '12 at 0:11
@Mic $\rm\ mod\ 5\!:\ 0^5\!\equiv 0,\ (\pm1)^5\!\equiv \pm1,\ (\pm2)^5\!\equiv \pm2.\ \ mod\ 6\!:\ m^5\equiv m^3 m^2 \equiv m\ m^2\equiv m\$ by OP. $\ \ \$ – Bill Dubuque Jul 17 '12 at 2:09
Bill, that is more clever than the solution I had in mind - I hadn't thought to exploit what was done in the original question! I was thinking of factoring as $m(m-1)(m+1)(m^2+1)$ and then arguing that at least one factor is divisible by each of 2, 3, and 5. – Michael Lugo Jul 17 '12 at 18:53

Lets start by factoring $n:$$n = m^3-m = m(m^2-1) = (m-1)m(m+1)$$ Note$(m-1),m$and$(m+1)$are three consecutive integers so (at least) one of these must be a multiple of$2$and one of these must be a multiple of$3$. - Thank you - that is helpful (I can probably do something now with factors) – George Jul 17 '12 at 0:16 Glad to help. Yes: $$n = (2p)\times (3r)\times t$$ where$p,r$and$t$are integers :) – Quixotic Jul 17 '12 at 0:17 so I did this:$\frac{n}{6} = ( (m - 1)(\frac{m}{2})(\frac{(m + 1)}{3}))$so$n = 6((m - 1)(\frac{m}{2})(\frac{(m + 1)}{3}))$is that cheating? Or am I going the right way? – George Jul 17 '12 at 0:31 I don't think this is the right way since if this is valid you can probably show divisibility by any number. – Quixotic Jul 17 '12 at 0:33 After what I said before,$n = (2p)\times (3r)\times t = 6 (prt)$and you are done. – Quixotic Jul 17 '12 at 0:42 Hint$\rm\ mod\ 6\!:\ 0^3\!\equiv 0,\ (\pm1)^3\!\equiv \pm1,\ (\pm2)^3\!\equiv \pm2,\ 3^3\!\equiv 3\:\Rightarrow\:n^3\equiv n\ \ $QED Note$ $It's easier via balanced residues$\{0,\, \pm1,\, \pm2,\, 3\}$vs.$\,\{0,1,2,3,4,5\},\,$by$\rm\:4\equiv -2,\:\rm 5\equiv -1.\:$It is not difficult to prove a generalized Euler-Fermat theorem, namely Theorem$\ $For naturals$\rm\: e,m,n\: $with$\rm\: e,m>1 \rm\qquad\qquad\ m\ |\ n^e-n\ $for all$\rm\:n\ \iff\ m\:$is squarefree and prime$\rm\: p\:|\:m\: \Rightarrow\: p\!-\!1\ |\ e\!-\!1 $Yours is the special case$\rm\:e={\bf\color{blue}3},\ m = 6 = {\bf\color{#C00}2}\cdot{\bf\color{#0A0}3}\:$is squarefree, and$\rm\, {\bf\color{#C00}2}\!-\!1,{\bf\color{#0A0}3}\!-\!1\:|\:{\bf\color{blue}3}-1.\$