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I'd like to know if my proof of the following is valid and if there's a shorter proof:

Prove that for all $n\in \mathbb{N}$ we have that $6$ divides $n^3-n$

Proof: We proceed by induction...

$P(1)=(1)^3-(1)=0$ which $6$ divides

Suppose the proposition is true for $n=k$, that is, there exists $j\in \mathbb{N}$ s.t. $k^3-k=6j$.

We show that $P(k+1)=(k+1)^3-(k+1)=6i$ for some $i$: $$(k+1)^3-(k+1)=k^3+3k^2+2k$$ $$=(k^3-k)+3k^2+3k$$ $$=\color{red}{6j+3k(k+1)}$$

Now we show that $6$ divides $3k(k+1)$. Either $3k(k+1)$ is even (case 1), or odd (case 2).

(case 1) There exists $l$ s.t. $k=2l\implies3k(k+1)=6l(2l+1)$

(case 2) There exists $m$ s.t. $k=2m+1\implies3k(k+1)=3(2m+1)(2m+1+1)=6(2m^2+3m+1)$

Now, (case 1) and (case 2) $\implies 6$ divides $3k(k+1)\implies 3k(k+1)=\color{blue}{6q}$ for some $q\in\mathbb{N}$

Next, $$P(k+1)=\color{red}{6j+3k(k+1)}=6j+\color{blue}{6q}=6(i)$$ Therefore, by induction $6$ divides $n^3-n$ for all $n\in \mathbb{N}$

Cotton Headed Ninnymuggins
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    +1: to your posting. Although it is true that easier proofs exist for this specific problem, the approach that you used is perfectly valid. Further, there may well be more complicated Number Theory problems where the approach that you have taken turns out to be the best approach. – user2661923 Oct 22 '22 at 22:33
  • A third (unmentioned) approach for problems of this type is to recognize that given any positive integers $r,s,t,n$ that $$r \equiv s \pmod{n} \implies r^t \equiv s^t \pmod{n}.$$ Therefore, for this type of problem, you could have simply manually checked each element in the set ${0,1,2,3,4,5}.$ – user2661923 Oct 22 '22 at 22:37

3 Answers3

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Your proof looks fine, but there's definitely a shorter proof: $$n^3-n=(n-1)(n)(n+1) $$ At least one of those three factors is divisible by $2$, and at least one is divisible by $3$, so their product is divisible by $6$.

Lee Mosher
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I think you're making things more complicated than needed. Just rewrite the given expression as follows: $$k^3 - k = k \times (k^2 - 1) = (k - 1) \times k \times (k + 1).$$ It is trivial that from n consecuitive numbers there is exactly one number dividable to n. So from the three consecutive numbers k - 1, k, and k + 1 there is exactly one number dividable to 3. And there is exactly one even number in the set {k, k + 1}. So $(k - 1) \times k \times (k + 1)$ is dividable to 2 and 3. Since 2 and 3 are relatively prime, $k^3 - k$ will be dividable to $2 \times 3 = 6$.

Saber Dinpashoh
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Since $n^3 \equiv n \pmod{2}$, we know that $2$ divides $n^3 - n$.

Now use induction to show that $n^3 - n$ is also divisible by $3$:

Base case: Certainly $3$ divides $n^3 - n$ when $n = 0$.

Inductive step:
Now suppose $3$ divides $n^3 - n$.
Algebra allows us to write

$\quad (n+1)^3 - (n+1) = n^3 + 3n^2 + 2n = n^3 - n + 3n^2 + 3n$

But then

$\quad [(n+1)^3 - (n+1)] \pmod{3}\equiv$
$\quad\quad [n^3 - n] \pmod{3} + [3n^2 + 3n]\pmod{3}\equiv$
$\quad\quad 0 \pmod{3} + 0 \pmod{3}\equiv 0\pmod{3}$

We conclude that $3$ also divides $n^3 -n$.

Since both $2$ and $3$ are prime factors, the product $6$ must also divide $n^3 -n$.

CopyPasteIt
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