Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
@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

3 Answers 3

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 $3$.

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

share|improve this answer

You could also show this by induction, noting $0^3-0$ is a multiple of $6$ for the base case, and then showing that $(m+1)^2-(m+1) - (m^3-m)$ is a multiple of $6$ for all $m$ for the inductive step. (It is enough to cover the nonnegative case, because $(-m)^3-(-m)=-(m^3-m)$.)

You can show that the new expression, which simplifies to $3m(m+1)$, is a multiple of $6$ either by noting that one of $m$ or $m+1$ is even as in Quixotic's answer, or again using induction. Note that $3\cdot 0(0+1)$ is a multiple of $6$ for the base case, and then show that $3(m+1)(m+2)-3m(m+1)=6(m+1)$ is a multiple of $6$ for the inductive step.

Applying a similar method to Michael Lugo's problem in the comments shows after $3$ steps of taking differences (and checking base cases) that $m^5-m$ is divisible by $30$. If $f(m)=m^5-m$, $g(m)=f(m+1)-f(m)$, $h(m)=g(m+1)-g(m)$, and $k(m)=h(m+1)-h(m)$, then $f(0)=g(0)=0$, $h(0)=30$, and $k(m)=150 + 180 m + 60 m^2$.

share|improve this answer
2  
I deleted this answer because it is excessively complicated. I thought it would be more helpful by not being here. It was undeleted by a moderator. –  Jonas Meyer Aug 7 at 3:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.