Use strong induction to prove that $12\mid n^4-n^2$ for every $n\in\mathbb{N}$

Question

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I am currently reading the Book of proof. There's an example problem on page 188, illustrating the method called Strong induction. The outline for proof by strong induction as follows:

Proposition The statements $S_{1}, S_{2}, S_{3}, S_{4}$,... are all true.

Proof. (Strong induction)

(1) Prove the first statement $S_{1}$. (Or the first several $S_{n}$, if needed.)

(2) Given any integer $k\ge{1}$, prove ($S_{1}\land S_{2}\land S_{3}\land S_{k})\implies S_{k+1}$.

The main difference between strong induction and normal mathematical induction is that is not only uses the item $S_{k}$ to prove that $S_{k+1}$, but it also uses the previous $m$ items ($m\le{k}$).

Then he uses strong induction to prove that $12\mid n^4-n^2$ for any $n\in\mathbb{Z}$.

The proof as follows:

1. Say $S_{n}: 12\mid n^4-n^2$
2. He prove that $S_{n}$ is true for $S_{1}$ to $S_{6}$
3. for $k\ge{6}$, assume $12\mid m^4-m^2$ for $1\le{m}\le{k}$.($S_{1}...S_{k}$ is true). We want to infer from $12\mid (k-5)^4-(k-5)^2$ to $12\mid (k+1)^4-(k+1)^2$. He set $k-5=l$ and $l^4-l^2=12a$，then $k+1=l+6$. Substitute $l+6$ into the $(k+1)^4-(k+1)^2$, and he finally reach $12(a+2l^3+18l^2+71l+105)$, successfully prove the proposition.

For the proof, he just said that:

In particular, if $S_{1}$ through $S_{k}$ are true, then $S_{k-5}$ is true, provided $k-5\ge{1}$. We will show $S_{k-5}\implies S_{k+1}$ instead of $S_{k}\implies S_{k+1}$. But as $k-5\ge{1}$, we have $k\ge{6}$. Then our basic steps must check that $S_{1}$ to $S_{6}$ are all true.

My question is: Why does he choose to use the first 6 items as the base to prove the following proposition. In other words, why choosing $S_{k-5}\implies S_{k+1}$ instead of $S_{k-2}\implies S_{k+1}$. I tried it, but failed with some coefficients not a multiple of 12. Are there any internal mathematical theorem here?

I don't know how to deal with the problem when I first came across this, and I saw him choosing the first 6 items as a group, and finally reached the conclusion that $n^4-n^2$ is a multiple of 12. His inference just looks like magic! I wonder if it comes from purely mathematical intuition or some intrinsic number theory theorem. There's no illustration on how he came across this idea from the textbook, so I am reaching out for help here.

Answer 1

Here $\:\!\color{#0af}6\:\!$ is chosen because it is some period of $\,S_n\bmod \color{#c00}{12},\,$ i.e. $\,\color{#c00}{S_{n+\color{#0af}6}\equiv S_n\pmod{\!4\cdot 3}},\,$ because $\:\!\color{#0af}6\:\!$ is common multiple of some periods $(\color{#0af}{2}\,$ and $\,\color{#0af}3)$ of $\,S_n\,$ taken $\!\bmod \color{#c00}4\,$ and $\! \bmod \color{#c00}3,\,$ resp.

Thus we can employ the following "$\!\!\bmod\color{#0af}6$" induction schema:

Lemma $\ $ If $\,\color{#00f}{S_1,S_2,\ldots S_{\:\!\color{#0af}6}}\,$ are true, and for all $\,k\!:\, \color{#c00}{S_k\Rightarrow S_{k+\color{#0af}6}}$ then $\,S_n\,$ is true for all $\,n\ge 1$.

Proof $\, $ (Base Cases) $ $ If $\,\color{#00f}{n\le \color{#0af}6}\,$ then $\,\color{#00f}{S_n}\,$ is true by hypothesis. (Induction Step) suppose $\,\color{#00f}n>\color{#0af}6\,$ and suppose for complete induction that $\,S_k\,$ is true for all $\,k < n.\,$ In particular $S_{n-\color{#0aaf}6}$ is true, which by $\rm\color{#c00}{hypothesis}$ implies that $S_n$ is true, so the claim follows by complete induction.

Remark $ $ Essentially the inductive step lifts the truth from any "remainder" index $\,k\le 6\,$ up to all naturals $\,n\equiv k\pmod{\!6},\,$ ie $\,k\to k+6\to k+12\to \cdots\,$ It reaches every natural because every natural $\,n\,$ is congruent to its remainder $\,k = n\bmod_+\! 6 \,\le\, 6,\,$ so the inductive lifting of truth from $\,k\,$ eventually reaches $\,n = k + 6j.\,$ Essentially the induction is "piggybacking" on the inductive proof of the division algorithm (here division by $\,6)$.