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

My attempt

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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    Both proofs seem correct. – ultralegend5385 Jun 26 '21 at 15:54
  • You can prove stronger result. $12|n^4-n^2$ – lone student Jun 26 '21 at 15:59
  • Please don't duplicate (your prior) questions. Verification questions require context, viz. precisely which part of the proof do you have doubt about, and why. – Bill Dubuque Jun 26 '21 at 20:08
  • Duplicated it because that question was closed and my edits were not being reviewed. I’m just making sure I’m understanding the material and whether my way of proving a proposition is correct. – Karam Jun 26 '21 at 22:00
  • If you don't provide the necessary context then your questions will likely be closed and deleted. The site is not a proof checking machine and should not be misused in that way. – Bill Dubuque Jun 26 '21 at 23:16

3 Answers3

1

You don't need induction, write $$n^4-n^2=n^2(n-1)(n+1)$$ a one line proof.

Dr. Sonnhard Graubner
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Your proof seems correct but messy.

There’s a much easier inductive style proof. You can show it for $0,1,2$ by calculation and then note that $3| (n+3)^4-(n+3)^2-n^4+n^2$ by basic modular arithmetic.

Eric
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Note that $n^4 - n^2 = (n^2 - n) (n^2 + n) = (n - 1) n^2 (n + 1)$, one of the three factors is divisible by $3$.

vonbrand
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