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:

case 1:

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)

case 2:

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 :)

Lawrence

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...?

>

> Adam

>

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