Proof by minimum counter example

Question

I need to prove that $n^4-n^2$ is divisible by 12 by minimum counter example. I understand the process but I don't understand how we arrive at m>=7. I have seen different proofs but I still don't know how to decide which value is the smallest that I can use.

Answer 1

$$(12-m)^{2r}=m^{2r}-2rm^{2r-1}12+\cdots+(12)^{2r}\equiv m^{2r}\pmod{12}$$ for integer $r>0$

$$\implies(12-m)^4-(12-m)^2\equiv m^4-m^2\pmod{12}$$

Answer 2

Expand the expression $(k+6)^4 - (k+6)^2$. The first power contributes $k^4$ plus additional terms (call these $A$) and the second term contributes $k^2$ plus additional terms (call these $B$). Now check whether $A$ and $B$ are perhaps divisible by $12$. If they are, then $k^4 - k^2$ is not divisible by 12, since you assumed that $m^4 - m^2$ is not divisible by 12. However, $k < m$, so now you have found a smaller counterexample.

Answer 3

They are basically showing (or telling you to show) that if $12\not\mid(m^4-m^2)$, then $12\not\mid (k^4-k^2)$, where $k=m-6$. This says that the minimal positive counterexample, if there is one, must be among the numbers $1\le m\le6$, since these are the only positive numbers for which $m-6$ is not positive (and thereby a smaller positive counterexample). That's why they start be showing that none of those numbers is a counterexample.