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Claim: $3\mid n^4-n^2$ for all $n \in \mathbb{Z^+}$, $n\ge2$.

What I've done so far:

Base case: Let $n=2$. Then $\exists k \in \mathbb{Z}$ such that $12=3k$, namely, $k=4$. Thus $3 \mid 12$ and hence the claim holds for $n=2$.

Assume for positive integer $m \ge 2$ that $m^4-m^2=3k'$ with $k' \in \mathbb{Z}$

I substituted $m+1$ for $n$, and have

$(m+1)^4 - (m+1)^2$ $=$ $\sum_{i=0}^{4} {4\choose i} m^i \cdot 1^{4-i} - (m+1)^2= m^4+4m^3+6m^2+4m+1-(m+1)^2=m^4-m^2+4m^3+6m^2+2m=(m^4-m^2)+4m^3+6m^2+2m=3k'+4m^3+6m^2+2m$

I'm stuck. Would this be the correct way to go?

Reattempt

Proposition: $3 \mid n^4-n^2$ for all $n \in \mathbb{Z^+}, n \ge 2$

Lemma: $3 \mid (m-1)m(m+1)$

Proof. Suppose $m \in \mathbb{Z}$. By the QRT, we have $m=3q+r$ $\,\ni\, r \in \{0,1,2\}$, and $q=\lfloor{\frac{m}{3}}\rfloor$.

We want to show that $(m-1)m(m+1)$ is divisible by 3.

Clearly, if any one of the three factors in $(m-1)m(m+1)$ is divisible by 3, then the product must also be divisible by 3.

$\cdot$ Case 1 ($r=0$): If $r=0$, then $m=3q$ is divisible by 3.

$\cdot$ Case 2 ($r=1$): If $r=1$, then $m-1=(3q+1)-1=3q$ is divisible by 3.

$\cdot$ Case 3 ($r=2$): If $r=2$, then $m+1=(3q+2)+1=(3q+3)=3(q+1)$ is divisible by 3.

In either case, one of the three factors in $(m-1)m(m+1)$ is divisible by 3. Thus $3 \mid (m-1)m(m+1)$. Therefore, the product of any 3 consecutive integers is divisible by 3.

Proof.

Base case: Let $n=2$. Then $\exists k \in \mathbb{Z}$ such that $12=3k$, namely, $k=4$. Thus $3 \mid 12$ and hence the claim holds for the base case.

Assume for positive integer $m \ge 2$ that $m^4-m^2=3k$ for some $k \in \mathbb{Z}$. We need only to show that this implies that $3 \mid (m+1)^4-(m+1)^2$.

$(m+1)^4-(m+1)^2= \sum_{i=0}^{4} {4\choose i} m^i \cdot 1^{4-i}-(m+1)^2=(m^4-m^2)+4m^3+6m^2+2m=(m^4-m^2)+3(m^3+2m^2+m)+(m^3-m)$.

By the inductive hypothesis, $3 \mid(m^4-m^2)$. And since $(m^3+2m^2+m) \in \mathbb{Z}$, it follows that $3 \mid 3(m^3+2m^2+m)$. Note that $(m^3-m)=m(m^2-1)=(m-1)m(m+1)$. By the Lemma, $3 \mid (m^3-m)$.

Thus we have $3 \mid (m^4-m^2)$, $3 \mid 3(m^3+2m^2+m)$, and $3 \mid (m^3-m)$.

To show that their sum is divisible by 3, let $k_1,k_2,k_3 \in \mathbb{Z} \,\ni\, m^4-m^2=3k_1, 3(m^3+2m^2+m)=3k_2$, and $(m^3-m)=3k_3$. Thus $3k_1+3k_2+3k_2=3(k_1+k_2+k_3)=m^4-m^2+3(m^3+2m^2+3m)+(m^3-m)$, where $(k_1+k_2+k_3) \in \mathbb{Z}$.

Hence $3 \mid (m+1)^4-(m+1)^2$.

Therefore, by induction, $3 \mid n^4-n^2, \forall n \in \mathbb{Z^+}, n \ge 2$

Karam
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2 Answers2

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If you want to continue this, then

you found $(m+1)^4 - (m+1)^2 = (m^4-m^2)+4m^3+6m^2+2m$

and this is $(m^4-m^2)+3(m^3+2m^2+m) +(m^3-m)$

so you have reduced the problem to showing $3|(n^3-n)$

and if you used induction for that you would have $(m+1)^3-(m+1) = (m^3-m) +3(m^2+m)$.

Reversing this, you could have a faster proof:

$$n^4-n^2 = 3n\sum\limits_{k=1}^{n-1} (k^2+k)$$ which is clearly divisible by $3$

Henry
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  • I don’t see how that implies “so you have reduced the problem to showing 3 divides (n^3 - n).” – Karam Jun 25 '21 at 13:32
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    @Karam in the expression $(m^4-m^2)+3(m^3+2m^2+m) +(m^3-m)$, you have $3| (m^4-m^2)$ as the inductive hypothesis and clearly $3| 3(m^3+2m^2+m)$ so that leaves the question of whether $3| (m^3-m)$ – Henry Jun 25 '21 at 13:35
  • Got it. But how would you use induction on that? Do we assume the hypothesis holds for m^3-m? So would that work as a lemma in this proof? – Karam Jun 25 '21 at 13:45
  • @Karam You can prove $3| (n^3-n)$ by induction using $(m+1)^3-(m+1) = (m^3-m) +3(m^2+m)$. You then say this implies $3| n (n^3-n)$ i.e. $3| (n^4-n^2)$ – Henry Jun 25 '21 at 14:46
  • I'm not sure I understand - are you relating $(n^3-n)$ to $(m+1)^3-(m+1)$? But how does $(m^3-m)+3(m^2+m)$ imply that $3| n (n^3-n)$? – Karam Jun 25 '21 at 18:56
  • I’m not sure how to use induction twice. – Karam Jun 25 '21 at 19:18
  • Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 25 '21 at 20:55
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    @BillDubuque Just out of curiosity, if the person who answers reacts to your message by deleting their own answer, is that just as good (for site maintenance) as having never posted the answer in the first place? – user2661923 Jun 26 '21 at 02:31
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Note that claim is

$3|n^4-n^2$

$\implies 3|n*(n-1)*n*(n+1)$

Clearly the last three terms written are three consecutive natural numbers , therefore at most one of them will be a multiple of three and at least one will be even , therefore overall it is divisible by 6

Now it is being multiplied by another natural number $n$ (the first term which we didn't consider in above para), therefore still it is a multiple of 6 and hence the whole number is divisible by 3.

In short , the place where you put $3k'$ in last step can be done earlier only.

Lalit Tolani
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