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I am not sure if I am doing this right, but I have this:

There exists an integer $k$. $2k =$ positive number

$(2k)^3 - 2k$ [*And this is where I get lost. How does one prove this?]

Bill Dubuque
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TheMedMajor
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    Let the "why induction, just note that (...)" comments/answers begin. – Git Gud May 10 '15 at 18:24
  • @GitGud Are you an oracle? ;) – Emily May 10 '15 at 18:37
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    Actually a proof by induction makes sense here. It is more evident that $f(k+1) - f(k)$ is divisible by three than that $f(k) = n^3 - n$ is. So check the first few base cases and let induction take over. – hardmath May 10 '15 at 18:41

6 Answers6

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Note that $n^3-n=n(n^2-1)=n(n+1)(n-1)$. Now we know that of any three consecutive numbers is at least one divisible by 3, so this gets your result.

wythagoras
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  • I didn't even think of breaking it apart like that. Thank you very much! – TheMedMajor May 10 '15 at 18:27
  • I don't mean to be rude, but since you are new to this site: If you like some answer, you might want to upvote it and click the accept button. Note that you can upvote all answers you want (even on other people's questions), but only accept one answer per question. – wythagoras May 10 '15 at 18:31
  • I cannot up vote till a certain time has elapsed. Or else I would have done so immediately. – TheMedMajor May 10 '15 at 18:33
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    @wythagoras You may not know that there is an embedded 15 minute delay between asking a question and accepting an answer. Rapid acceptance is generally discouraged, as it precludes other alternative answers. – Emily May 10 '15 at 18:33
  • Okay, never mind then, and sorry... – wythagoras May 10 '15 at 18:35
  • This answer deserves these remarks. A very simple and great answer. – Sufyan Naeem May 10 '15 at 18:42
  • All answers given here are wonderfully written and thought of. I just want to thank everyone again! – TheMedMajor May 10 '15 at 18:45
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For $n=1$, the result is trivial, so let's use $n=2$ as our base case: $2^3-2 = 8-2 = 6 \equiv 0 \bmod 3$.

Now, to use induction, we assume that it is true for $n$, and we want to prove that this must imply that it is true for $n+1$.

Let $3 \mid n^3-n$.

Now, $$\begin{align*} (n+1)^3-(n+1) &= n^3+3n^2+3n+1-(n+1) \\ &= n^3+3n^2+2n \\ &= \color{red}{n^3 -n} + \color{blue}{3n^2 +3n}. \end{align*}$$

Now, by our induction assumption, the term in red is divisible by $3$, and the term in blue is obviously divisible by $3$.

Therefore, by induction, $n^3-n$ is divisible by 3.

Emily
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  • Very clean and neat! Thank you Arkamis. – TheMedMajor May 10 '15 at 18:42
  • In particular, to say that result is so trivial for the base number and take a number greater than it, make the induction proof weaker. Better to check for $1$ first. Our sense instantly accept it but Math always need a trial and error method. – Sufyan Naeem May 10 '15 at 18:52
  • @SufyanNaeem I am giving the OP enough credit to recognize that $1-1= 0$ is trivially true. – Emily May 10 '15 at 18:55
  • It is not the place to award some one with credit if he/she pays a lot of attention. The answer should be as much improved as it can be and tends to be general. – Sufyan Naeem May 10 '15 at 18:59
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Let the above be a proposition denoted by $P_n$.

Base case:

Let $n=1$,

$1^3-1=0$ and since $0$ is divisible by $3$, $P_n$ holds for $n=1$.

Inductive step:

Let the proposition be true for $n=k$ given by,

$k^3-k$ is divisible by $3$

Then it holds for $n=k+1$.

$(k+1)^3-(k+1)$

=$k^3-k+3(k^2+k)$ We see that $3(k^2+k)$ is divisible by $3$ and since we assumed that $k^3-k$ is divisible by $3$ therefore it is proved that $P_n$ holds $\forall n \in \mathbb N$

Sufyan Naeem
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3

Below are a handful of possible approaches.

First, notice $\displaystyle \ \ n^3\!-n = n(n^2\!-1) = (n\!+\!1)n(n\!-\!1) = \color{#c00}6{ {n\!+\!1\choose 3}} \ $ is divisible by $\,\color{#c00}6$

Alternatively show $\,(n\!+\!1)n(n\!-\!1)\,$ is divisible by both $2$ and $3$. Generally a sequence of $\,k\,$ consecutive integers contains a multiple of $\,k\,$ (hint: let pigeonholes be remainders mod $\,k).$

More generally $\ 2p\mid n^p-n\ $ for odd primes $p,\,$ since $\ p\mid n^p-n\ $ by little Fermat, and it is divisible by $2$ since $\,n^p$ and $\,n\,$ have equal parity. Even more generally

Theorem $ $ (Korselt's Pseudoprime Criterion) $\ $ For $\rm\:1 < e,m\in \Bbb N\:$ we have $$\rm \forall\, n\in\Bbb Z\!:\ m\mid n^e\!-n\ \iff\ m\ \ is\ \ squarefree,\ \ and \ \ p\!-\!1\mid e\!-\!1\ \, for\ all \ primes\ \ p\mid m$$

For a proof see this answer. Your case is $\,\rm\,e,m = 3,6\,$ and, indeed, $2\!-\!1,\,3\!-\!1\mid 3\!-\!1.$

Community
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Bill Dubuque
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Without using induction you can note that $$n\cdot n^2\equiv n\cdot 1 \pmod 3$$ (using the Little Fermat's theorem and $n\neq 3k$) therefore $n^3-n\equiv 0 \pmod 3$

wythagoras
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Domenico Vuono
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We could also play it modular

$$n\equiv 0,1,2\pmod{3}$$ $$-n\equiv 0,2,1\pmod{3}$$ $$n^3\equiv 0,1,2\pmod{3}$$

And we can see that we always have

$$n^3-n\equiv 0\pmod{3}$$

marwalix
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