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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

2 Answers 2

up vote 4 down vote accepted

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 $3$.

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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 $\rm\, {\bf\color{#C00}2}\!-\!1,{\bf\color{#0A0}3}\!-\!1\:|\:{\bf\color{blue}3}-1.$

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