Prove that $\displaystyle 12\ |\ n^2-1$ if g.c.d. (n,6) = 1.

Printable View

- Oct 19th 2009, 04:07 PMkyldn6Fermat's Little Theorem
Prove that $\displaystyle 12\ |\ n^2-1$ if g.c.d. (n,6) = 1.

- Oct 19th 2009, 07:32 PMchisigma
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...

$\displaystyle 2\cdot 2\cdot 3 \mid (n+1)\cdot (n-1)$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 19th 2009, 09:47 PMtonio