dwheatley
Joined: Nov 16, 2009
• Posts: 1246
December 21st, 2013 at 1:08:01 PM permalink
Quote: Greasyjohn

I completely understand that there's no largest prime. I was merely trying to continue the calculation. Perhaps there is no calculation--multiply any know primes and add one and there' s always a prime greater than ANY prime on our list.

Right. Do the multiplication, add one, and either

a) the result will be prime, or
b) a prime larger than one in the list will divide the result.

Maybe you can characterize when this will happen and will a Fields medal? ;)
Wisdom is the quality that keeps you out of situations where you would otherwise need it
Greasyjohn
Joined: Dec 8, 2013
• Posts: 2169
December 21st, 2013 at 1:08:10 PM permalink
Quote: Twirdman

Yes if you take a list of all the primes up to a certain value say 23 so you have 2*3*5*7*11*13*17*19*23+1 you will get a larger prime. Though you have to still find what that larger prime is by doing repeated divisions of the number and there are simply much easier ways to find incredibly large primes.

But you could have any primes in your calculation? Is 3x17x509 just as valid as any more complete list of primes?
dwheatley
Joined: Nov 16, 2009
• Posts: 1246
December 21st, 2013 at 1:09:22 PM permalink
If you skip primes the result will be either:

a) the number is prime, or
b) a prime not in your list divides the result.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
Twirdman
Joined: Jun 5, 2013
• Posts: 1004
December 21st, 2013 at 1:14:32 PM permalink
Quote: Greasyjohn

But you could have any primes in your calculation? Is 3x17x509 just as valid as any more complete list of primes?

Yeah that is a valid list but you won't necessarily get a bigger prime then just one not on the list. Like I showed with 3*5+1=16 that is 2^4 so 2 is the new prime you got but its not bigger than 3 or 5 its just not on your list.
AxiomOfChoice
Joined: Sep 12, 2012
• Posts: 5761
December 21st, 2013 at 1:14:45 PM permalink
Quote: Twirdman

Yes if you take a list of all the primes up to a certain value say 23 so you have 2*3*5*7*11*13*17*19*23+1 you will get a larger prime.

No, that's incorrect. There's already been a counter-example given in this thread which shows that that's wrong (2*3*5*7*11*13 + 1 = 59 * 509)
Twirdman
Joined: Jun 5, 2013
• Posts: 1004
December 21st, 2013 at 1:28:31 PM permalink
Quote: AxiomOfChoice

No, that's incorrect. There's already been a counter-example given in this thread which shows that that's wrong (2*3*5*7*11*13 + 1 = 59 * 509)

And 59 is a larger prime then any on the list. I could have worded it better I guess but I was just saying it will prove the existence of a larger prime which you can find by either a simplistic algorithm of just dividing or some more advanced algorithms that I don't know since my area of math isn't number theory or algorithms.
Greasyjohn
Joined: Dec 8, 2013
• Posts: 2169
December 21st, 2013 at 1:32:57 PM permalink
Quote: AxiomOfChoice

No, that's incorrect. There's already been a counter-example given in this thread which shows that that's wrong (2*3*5*7*11*13 + 1 = 59 * 509)

I'm guessing that Twirdman means not "will get a larger prime" but that there IS a larger prime than is on his list.
Twirdman
Joined: Jun 5, 2013
• Posts: 1004
December 21st, 2013 at 1:35:26 PM permalink
Quote: Greasyjohn

I'm guessing that Twirdman means not "will get a larger prime" but that there IS a larger prime than is on his list.

Yeah I will admit I worded that badly and should have been more careful. I normally am but think I got lazy since I'm on break.
98Clubs
Joined: Jun 3, 2010
• Posts: 1728
December 24th, 2013 at 8:14:57 PM permalink
One of the ways to attack the problem of large primes is a rather simple rule that will eliminate 60% of all numbers larger than 5.

In a rather odd way, except for the number 2 and the number 5, all known primes have a final digit of either 1, 3, 7, or 9. However further testing is still needed: 21 is not prime as is 49 for example.
Some people need to reimagine their thinking.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4758
December 25th, 2013 at 10:37:42 AM permalink
Quote: 98Clubs

One of the ways to attack the problem of large primes is a rather simple rule that will eliminate 60% of all numbers larger than 5.

In a rather odd way, except for the number 2 and the number 5, all known primes have a final digit of either 1, 3, 7, or 9. However further testing is still needed: 21 is not prime as is 49 for example.

I'll do better than that; I will eliminate 2/3 of all numbers greater than 3.

If n > 3 and is prime, then either n+1 is a multiple of 6 or n-1 is a multiple of 6.
Proof: if n / 6 has a remainder of 0, 2, or 4, then it is a multiple of 2; if its remainder is 3, it is a multiple of 3; if it is 1, then n-1 is a multiple of 6; if it is 5, then n+1 is a multiple of 6.