Expression of Prime numbers via $ak\pm b$ This is extension of my previous question 
How to prove Prime numbers can be expressed as $6k\pm 1$
The main answer from the post is list all $6k\pm1, 6k \pm2, 6k +3, $ and see wheter it can be decompose into other numbers, $i.e$, $2$ or $3$. 
At this stage, i have a question why $6$?  
How about 4, 5, 7, or any other numbers and forms like $ak+b$?
Is there any reason for $6$? 
 A: To have $ak \pm m $ describe all primes, then $m $ and $a-m $ must be the only two numbers smaller than $a$ that are relatively prime to $a $.
For $2$, $2k+1$ covers all primes (except 2) because the only other cases are $2k $ and those are composite (except 2).
Same with $3k\pm 1$.  The only other case is $3k $.
$4k\pm 1$ describes all primes (except 2) because the only other case is $4k$ and $4k +2$.  But $2$ is not relatively prime to $4$ so all $4k+2$ will be composite.
We can't do anything with $5k \pm i $.  $5k$ must be composite, but as $1,2,3,4$ are all relatively prime to $5$ we can't say any of the $5k+i $ must be composite.  $5k+i $ could be prime for all $1,2,3,4$.
$6k \pm 1$ describe all primes because none of $i=0,2,3,4$ are relatively prime to $6$ so all $6k +i $ are composite.
We can't do anything with $7k \pm m $ because all $i=1,2,3,4,5,6$ are realitively prime so none of $7k+i $ must be composite.
For $8$ $8k\pm 1$ and $8k\pm 3$ describe all primes (except 2) because all other $i=0,2,4,6$ are not relatively prime to $8$ so $8k+i $ must be composite.  But $\pm 1$ and $\pm 3$ are both relatively prime so we need both of them and we can't make this a single expression.
And for no other number higher than $6$, can we have only one pair of relatively prime numbers.  So $6k\pm1$ is the last such single expression that describes all primes.
