For all $(n\in\Bbb N*)(\exists k \in\Bbb N)$ $ show $ : $n²(n²-1)=12k $

Question

We had this this exercise on the exam and I unfortunately I couldn't do it right but at least I tried. Here is what I did :

$$ (\forall n\in\Bbb N*)(\exists k \in\Bbb N) : n^2(n^2-1)=12k$$ The first thing I did was Disjunction of cases If $n$ is pair then : $$ n=2k$$ $$ n²=4k²$$ $$ n²(n²-1)= 4k²(4k²-1)$$ $$ n²(n²-1)= 16k^4-4k²$$ and that's where I got stuck So I thought I would try Proof by Induction For $n=1$ $$ 1²(1²-1) = 0 $$ $0$ is divisible by $12$ Which means the proposition is right for $n=1$ Let's suppose that : $$ n^2(n^2-1)=12k$$

Let's prove that : $$ (n+1)^2((n+1)^2-1)=12h$$ $$ (n+1)^2((n+1)^2-1) = (n+1)^2(n^2+2n)$$ $$ =(n^2+2n+1)(n^2+2n)$$ $$ = n^4+2n^3+2n^3+4n^2+n^2+2n $$ $$ = n^4+4n^3 +5n^2 +2n $$ And this is where I got stuck again. I know we already passed the test but I want to know the answer. At least I tried, please don't down vote.

Answer 1

You only have to see that $n^2(n^2-1)$ is always divisible by $4$ and $3$ (because $4$ and $3$ are relative primes).

To check the $3$, only notice that $$n^2(n^2-1)=n\cdot n(n-1)(n+1)$$

so you have three consecutive numbers, that means that at least one will be divisible by $3$

To check the divisibility by $4$, you have two cases.

• If $n$ is even, $n=2k$, so $n^2=4k^2$, letting your factor $4$.

• If $n$ is odd, $n=2k+1$, so $n^2=4k^2+4k+1$, so $n^2-1=4(k^2+k)$, letting your factor $4$ too.

Answer 2

Hints: $$ 2^2\mid n^2(n-1)(n+1). $$ and $$ 3\mid n^2(n-1)(n+1). $$

Answer 3

It is much more insightful to prove a slight generalization (OP is special case $\,p=3)$

Theorem $\ $ If $\,p\,$ is an odd prime then $\,4p\mid n^{\large 2}(n^{\large p-1}-1) =: a$

Proof $\ $ If $\,2\mid n\,$ then $\,4\mid n^{\large 2}.\,$ Else $\,n\,$ is odd so $\,{\rm mod}\ 4\!:\ n\equiv \pm1\,\Rightarrow\, \color{#c00}{n^{\large 2}\equiv 1}\,$ therefore $\ n^{\large p-1}\! = (\color{#c00}{n^{\large 2}})^{\large (p-1)/2}\equiv \color{#c00}1^{\large (p-1)/2}\equiv 1\,$ so $\,4\mid n^{\large p-1}-1.\ $ Thus in all cases $\,4\mid a.\ $

By litlle Fermat $\,p\mid n^{\large p}-n \ $ so $\ p\mid a = n(n^{\large p}-n).\ $

Since both $\,4,p\mid a\,$ so does their lcm = product $= 4p.\ \ $ QED

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Remark $\ $ A similar proof leads to the following generalization of the Euler-Fermat Theorem.

Theorem $\ $ Suppose that $\rm\ n\in \mathbb N\ $ has the prime factorization $\rm\:n = p_1^{e_{\:1}}\cdots\:p_k^{e_k}\ $ and suppose that for all $\rm\,i,\,$ $\rm\ e\ge e_i\ $ and $\rm\ \phi(p_i^{e_{\:i}})\mid f.\ $ Then $\rm\ n\mid (ab)^e\,(a^f-b^f)\ $ for all $\rm\: a,b\in \mathbb Z.$

Proof $\ $ See this answer.