Thread: Prime Number Theory

I have to show that every prime can be expressed in the form with an integer.

So I figure I need to show that is always an integer. If the numerator is divisible by 24 then it must be divisible by i.e. divisible by 2 at least three times. How can I show this is true?

My idea is to show that is always divisible by 8 and 3 but I don't know how to do it. I'm thinking something about the last digit of only taking certain values meaning that is divisible by 8 but am having difficulty deciding how to do it. Maybe I'm going about this the wrong way.

Any help is appreciated.

the primes are odds after 5.

x^2 = (x-1)(x+1) +1

so one less than a prime number will be divisible by (x-1) and (x+1)

the two even numbers above and below the prime number will both be divisible by 2, and one of them must be divisible by 4 as well. Show this will give you the divisibility by 8 that you want.

also, here's a hint on divisibility by 3.

pick any three consecutive integers. how many of them are divisible by 3? how can you show this? how is this useful to what you are trying to prove?

Since p is a prime, either (p-1) or (p+1) is divisible by 3.
So, (p-1)(p+1) is divisible by 3.----(1)

Since p is a prime, one of the factors {ie (p-1) or (p+1)} is divisible by 4 and the other is divisible by 2.
So, (p-1)(p+1) is divisible by 8.-----(2)

From (1) and (2), (p-1)(p+1) is divisible by 24.

It is easy to demonstrate that the relation holds if p is any odd number not multiple of 3...

Kind regards

It is trivial . What u have stated is nothing but stating that all prime numbers other than 3 and 5 are of the form 6k+1 or 6k-1.
If p =6k+1 then p^2 -1 = 36k^2 + 12k . This is divisible by 24 .
First divide by 12 then u will have 3 k^2 + k . This is always even thus divisible by 2 . hence p^2 -1 is divisible by 24.
Same way consider p =6k-1 .
I guess you can verify it from here