Prove that there are infinitely many primes of the form 6x-1.

I have been fighting this proof for an hour. I have no idea how to get started or on how to finish. Any help would be greatly appreciated.

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- Apr 10th 2008, 12:32 PMpadsinseveninfinite primes
Prove that there are infinitely many primes of the form 6x-1.

I have been fighting this proof for an hour. I have no idea how to get started or on how to finish. Any help would be greatly appreciated. - Apr 10th 2008, 01:19 PMAryth
All integers can be represented as:

Where:

and

Now, first we try to find the statements for which this is seemingly prime:

It's immediately clear that this is divisible by 6.

This has no immediate factors, MAY be prime.

It is immediately clear that it is divisible by two.

It is immediately clear that it is divisible by three.

Immediately clear that it is divisible by two.

It has no immediate factors. MAY be prime.

The only candidates for primacy are:

Or

But, for

Therefore, all integers that are candidates for primacy are:

Since it has already been proven that there are infinitely many primes, then there are also infinitely many primes represented by each:

and

Thus, there are infinitely many primes represented by

I know it may not be as good as you need. But you can use it to build your own proof if you would like. - Apr 16th 2008, 12:18 PMPaulRS
Excuse me, but this part is wrong, you have not shown that there are infinitely many primes of the form . Maybe there are infinetely many prime numbers of the form and a finite number of primes of the form (Evilgrin)

Indeed, suppose there's a finite number of primes of the form , let be all the primes of that form

Let

This number must have a prime factor of the form , but it is not divisible by any prime of the form (by our assumption). This is absurd, therefore there have to be infnitely prime numbers of the form - Apr 16th 2008, 03:48 PMThePerfectHacker
We can generalize this result. Let . Suppose that all the (odd) primes are either . Then there are infinitely many primes . And the proof is similar. Suppose there are finitely many and define . Now any prime divisior of has form . It cannot be that all have form because then would have that form too. So it must mean that one of the prime factors of satisfies but then the LHS is divisible by but not the RHS.

- Nov 3rd 2008, 09:18 PMr0r0trog
- Jan 22nd 2009, 09:39 AMstar_tenshiQuestion
I have a similar type of problem on an assignment, but with instead. However, I'm not allowed to use modular arithmetic. Using the example, I was just wondering if I could restate this last part as:

By the Fundamental Theorem of Arithmetic, there must be a prime of form such that . But is not divisible by any prime of form by our assumption. This is a contradiction, thus, there must exist infinitely many primes of the form .

Such a proof should also hold for the case of primes in the form of too right? - Jan 22nd 2009, 10:03 AMThePerfectHacker
That is what post #4 says.

All odd primes are either: .

If there are finitely many such primes of form it means we can form .

This number factors into odd primes. Say all these primes where of form . Then would be of form which is a contradiction. Thus, there exists , and odd prime divisor of form . But that forces which is a contradiction. Thus, there are infinitely many such primes.

And the same argument works for primes of form also.

Note: In case you are wondering proving that there are infinitely many primes of forms is more difficult. - Jan 22nd 2009, 10:29 AMstar_tenshi
Awesome, thanks!

- Jan 30th 2009, 10:42 AMWhoever
Would this work?

Numbers at 6n-1 are prime unless one of their factors is at 6n-1. If x is the highest prime at 6n-1 then every number > x at 6n-1 has a product at 6n-1 < x. It must be easy to show that this is impossible.