Re: [PrimeNumbers] Extension of Dirchlet Theorem
- I dont think any exist besides a few trivial cases.
First, p must be congruent to either 1 or 2 mod 3 because if it was
congruent to 0 it would be a multiple of 3.
so we have two cases:
p = 1 mod 3
2p-1 = 1 mod 3
4p-1 = 3 mod 3 = 0, so 4p-1 is a multiple of 3 (not prime)
p = 2 mod 3
2p-1 = 3 mod 3 = 0 so 2p -1 is a multiple of 3 (not prime)
4p-1 = 5 mod 3 = 2
thus, this never works unless we take 3 to be that multiple.
so this implies in case 1, p = 1 (not a solution)
or in case 2, 2p - 1 = 3, p = 2
so p = 2, 2p-1 = 3, 4p-1 = 7 so (2,3,7) is a solution
lastly, p could be = 0 mod 3, so p = 3, 2p-1 = 5, 4p-1 = 11
so (3,5,11) is also a solution.
Thus there are only two solutions and your search is futile :)
On Sat, 1 Nov 2003, Adam wrote:
> RE: Adam Karasek post for pythagorean triples with a and c primes and
> b a product of at almost four primes, I seek primes p with 2p-1 and
> 4p-1 simlutaneously prime (to reformulate, we could write p=2n+1, and
> then seek 2n+1,4n+1,8n+1 simultaneously prime). I am having poor
> luck finding such p values, contrary to the belief in the extension
> of Dirichlet's theorem. For instance, I generate a random 12 digit
> prime, search the through the next 5000 primes, and find about 200-
> 300 that have 2p-1 prime, but only 0 of those having 4p-1 prime.
> Am I not being patient enough or...?
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