Prove that $\displaystyle 12\ |\ n^2-1$ if g.c.d. (n,6) = 1.
If g.c.d. (n,6) = 1 that means that $\displaystyle 2 \nmid n$ and $\displaystyle 3 \nmid n$, i.e. n is and odd number non multiple of 3. But if $\displaystyle 2 \nmid n$ and $\displaystyle 3 \nmid n$ means also that $\displaystyle n-1$ and $\displaystyle n+1$ are both even numbers and 3 devides one of them. The conclusion is that...
$\displaystyle 2\cdot 2\cdot 3 \mid (n+1)\cdot (n-1)$
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$\displaystyle \chi$ $\displaystyle \sigma$