1. Jan 11, 2013

### cragar

1. The problem statement, all variables and given/known data
Except 2 and 3 , prove that their an infinite amount of primes
of the form 6m+1 and 6m+5 for some integer m
It says to use Euclid's method but replace the +1 with a -1.
3. The attempt at a solution
Would I just multiply some of these forms together and subtract 1
$(6m+1)....(6n+1)-1=x$
If I divide my new number x by (6m+1) it wont divide it evenly.
this doesn't seems like it proves it, am I on the right track?

2. Jan 11, 2013

### haruspex

I wouldn't think so. You would only be able to show that x is not divisible by primes of the form 6m+1. You won't be able to show it is not divisible by other primes.

3. Jan 11, 2013

### Staff: Mentor

If you multiply k primes of the form (6m+1) and l primes of the form (6m+5), you get a product of the form (6m+1) for even l and (6m+5) for odd l. This allows to find new primes, assuming a finite number of primes of those types. I am not sure how to get an infinite number of (6m+1)-primes, but it is certainly possible for (6m+5)-primes.

4. Jan 11, 2013

### cragar

ok thanks for the replies Ill work on it.

5. Jan 11, 2013

### haruspex

Suppose p to be the largest prime of a given one of those two types, and construct the product of all primes (except a certain few) up to p.

6. Jan 14, 2013

### cragar

So I make a product of all primes of the form 6m+1 and then add one
to it. Im still not really sure exactly what you guys are telling me, Ill keep thinking about it.
another thing i noticed is that all of the odd numbers greater than 5 are of the form
6m+1 or 6m+3 or 6m+5, but we know that 6m+3=3(2m+1) so it never produces primes.
so we know because their are an infinite amount of odd primes that
either or both 6m+1 or 6m+3 contain an infinite amount of primes.

Last edited: Jan 14, 2013
7. Jan 15, 2013

### haruspex

That's not what I wrote: "the product of all primes (except a certain few) up to p."

8. Jan 15, 2013

### Staff: Mentor

1 is a bad thing to add, as the result will be divisible by 2.
Right