Let me see if I have understood your question.
Well, if p is greater than 2 then it will not be divisible by 2. It has to be odd. There must be a remainder when we divide by 2. And this remainder has to be 1.
So
So:
So p-1 is divisible by 2.
More
So: Each prime p>2, is surrounded by two even numbers.
Now consider 3 as a divisor. Each prime p > 3 can not be divisible by 3. There must be a (nonzero) remainder.
This reminder is either 1 or 2.
So, either
So either
So either p-1 or p+1 is divisible by 3.
But both p-1 and p+1 are divisible by 2.
So either p-1 or p+1 is divisible by 6.
MORE: consider 4: p-1 and p+1 are both even and p+1 = (p-1)+2, so one of these two numbers is divisible by 4. (the multiples of four are every other multiple of 2: 2,*4,6,*8,10,*12,...)
SO: each prime has a multiple of 4 on one side or the other, as well as a multiple of 6.
So (visualize the numbers in line with n*q+1, n*q+2, ....n*q+(n-1) sliding below them...)
(I guess) Example: each prime>14 is within 4 of a multiple of 7.
That's about as far as I can go.... I think more complex patterns need some complicated Number Theory.
I hope this helps.... don't lose the interest in numbers. They are an endless comfort and distraction... and an occasional source of fun.
. . . . . . . . . ( end of section A question about Prime Numbers) <<Contents | End>>