Prove that there are infinitely many primes of the form 3n + 1.
Assume n is odd. Then 3n is odd and 3n+1 is even, so cannot be prime. So n is even, and proof that there are infinitely many primes of the form 6n + 1 is equivalent to the given problem.
Note this table:
(6k1)(6l1) = 6Q+1
(6k1)(6l+1) = 6Q1
(6k+1)(6l+1) = 6Q+1
where (6k+/1)and (6l+/1) are primes, and 6Q+/1 is their product.
Now assume that the problem required proof of infinitely many primes of the form 6k  1, then we would be nearly done, since following Math Man's method, we can take the product of all the primes of form (6k1), on the assumption that they are finite in number, and the result is still 6Q1. Accordingly, the number 6(6Q1)1 is either prime itself, i.e a new and larger prime of that form, or must have an odd number of prime factors of that form; see the table, none of which can be in our finite list. So there are infinitely many primes of the form 6k  1.
Now take the product of all the primes of form (6k+1), q, on the assumption that they are finite in number.
Since q is odd, let q=(2n+1), so that q^2 = 4n^2+4n+1.
Add 3: q^2+3 = 4n^2+4n+4 = 4(n^2+n+1).
Note next that (n^2+n+1)(n1) = n^31. Assume that a prime, p, divides n^31. If so, Fermat comes to the rescue, and either p divides 3 (true only if p=3), or p=3k+1, but as discussed above, this must mean p=6k+1. But if every prime factor of n^31 is of that form, the same must be true of its own factor (n^2+n+1).
So we can plug that back into what we had before: q^2+3 = 4(n^2+n+1), where every prime factor of n^2+n+1 is either 3 or a prime of form 6k+1. Since we eliminated every known prime of that form by adding 3 to q^2, each such prime factor must be a previously unidentified one.
Hence, there are infinitely many primes of the form 3n + 1.
Edited on July 4, 2015, 6:36 am

Posted by broll
on 20150704 06:28:58 