... because it would probably then indicate some sort of pattern in the generation of the list of primes...
I thought about this problem a bit cause I think some one is also offering a million to solve this and I think I would prove it as follows (avioding the pattern in primes): 1)(this is the catch) get some kind of lower bound on the number of primes lower than a given even number. 2)Now, what we want is say our even number is m we want a prime p such that m-p is also prime. So first throw out all the even numbers and pair up the rest of the numbers less than m with thier 'm-p partners'. 3) Now we have say n primes and we can put one and only one into each of the [m/4] pairs so we if we have more than [m/4] primes two must go in the same slot (pigeon hole priciple) and blamo, we're done.
Well I guess we've reduced it to showing that the number of primes less than a given number, , must be greater than [m/4].
Slashdot Mirror
... because it would probably then indicate some sort of pattern in the generation of the list of primes ...
I thought about this problem a bit cause I think some one is also offering a million to solve this and I think I would prove it as follows (avioding the pattern in primes):
1)(this is the catch) get some kind of lower bound on the number of primes lower than a given even number.
2)Now, what we want is say our even number is m we want a prime p such that m-p is also prime. So first throw out all the even numbers and pair up the rest of the numbers less than m with thier 'm-p partners'.
3) Now we have say n primes and we can put one and only one into each of the [m/4] pairs so we if we have more than [m/4] primes two must go in the same slot (pigeon hole priciple) and blamo, we're done.
Well I guess we've reduced it to showing that the number of primes less than a given number, , must be greater than [m/4].