Are all the primes greater than $3$ in form of $6n\pm1$ today I've encountered the following excercise:
Show that all the primes greater than $3$ are in form of $6n\pm1$...
My initial thought was that numbers are in form of $6n,\quad 6n\pm1,\quad 6n\pm2, \quad 6n\pm3$ in respect to $6$ and $6n\pm2=2(3n\pm1),\quad 6n\pm3=3(2n\pm1)$ so since they are multiples of $2$ and $3$ they can't represent primes. This however didn't completely satisfy me, because I've shown that other forms in respect to $6$ aren't primes, but that doesn't necessarily mean that $6n\pm1$ must be prime. I thought perhaps I can do induction but I don't know what inital property to use for a start, or perhaps I am overseeing it right now.
Thanks in advance!

It's because $6n+3$ is divisible by $3$
(and obviously, $6n$, $6n+2$, and $6n+4$ are all divisible by $2$)
So any prime must be of the form $6n+1$ or $6n+5$ (i.e. $6n-1$) ... which of course does not mean that all such numbers are primes .. but that wasn't the question.
