You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 147

I’ve got two ideas which if proven wrong will hopefully be a learning experience.

The first is a different way to do the Fermat Primality test.

And the second is that 2p^2 + 1 if it passes the Fermat Primality test and is not factorable by 3 will be prime.

Anyone interested? (I don’t know anything about Maths really, apart from GCSE and A-levels and my own research)

**"Time not important. Only life important."*** - The Fifth Element 1997*

Offline

**Grantingriver****Member**- Registered: 2016-02-01
- Posts: 122

The second proposition can be proven or disproven as follows:

If we assume that p is a prime number then we would have:

but we know that p is a prime hence the last equation is impossible except for the prime number 3 and hence:

Therefore the only integer which is a prime and satisfies both the conditions, stated in the proposition, is the prime 3. However, if we assume that p is not prime we should exclude all of the integers which are not a multiple of 3 for the same argument stated above. So this will produce the following expression which satisfies the second condition of the proposition (that the number is not factorable by 3):

Testing the n values from 1 to 40 did not produce any pseudoprime that satisfies the second condition (the Fermat primality test). However, if a value of the latter expression at any given integer n turns out to be pseudoprime then it will provide a counterexample that disproves the proposition. On the other hand, if it is proven to have no pseudoprime up to a particular value of n then the previous argument constitutes a proof of the proposition for all integers below the value of the expression corresponding to the related value of n.

Offline

**Primenumbers****Member**- Registered: 2013-01-22
- Posts: 147

Thanks a lot Grantingriver! I think I understood all that you said. That’s great stuff.

I’ve actually posted on another forum where the idea’s been changed to:

**If an integer, 2p + 1, where p is a prime number, is a divisor of the Mersenne number **

My argument is that because divisors of the Mersenne number

can’t be < p if p is a prime number. Therefore if 2p +1 is a divisor of it has no divisors as p is > the square root of 2p + 1. This will therefore make 2p + 1 a prime number.What I’m also interested in is how to put the idea that:

** An integer minus one, divided by 2, p times equals zero, will be a factor of **

**"Time not important. Only life important."*** - The Fifth Element 1997*

Offline

Pages: **1**