20081114, 15:18  #1 
Oct 2007
2×17 Posts 
"Variations on a theme by Euclid"
Here's an interesting proof that primes go on forever.
( 3*5*7*11*...b) + q =2^n On the left side of the equation you have sequence of primes ending in prime b which added to an odd number q . If the squence of primes and q have a common factor then 2^n would have to be evenly divisable by the common factor which it cannot be, therefore q must be prime or be made up of factors greater than b. This proof clearly indicates that primes go on forever. 
20081114, 15:55  #2 
Nov 2008
2×3^{3}×43 Posts 
You have to make it clear that 2 is not included in the sequence of primes.
Otherwise, good proof! P.S. Are you sure noone has thought of it before? I've never seen it before. Last fiddled with by 10metreh on 20081114 at 15:56 
20081114, 16:56  #3 
Oct 2007
42_{8} Posts 
This definitely a new proof unless I accidently reproduced somsone elses
work I've never seen before. 
20081114, 17:07  #4 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010101011_{2} Posts 
I'm not sure, but it sounds to me like a roundabout version of Euclid's proof of infinite primes.
http://en.wikipedia.org/wiki/Prime_n..._prime_numbers Last fiddled with by MiniGeek on 20081114 at 17:47 Reason: Euler, Euclid, what's the difference? both old guys with "Eu" in their names...;) 
20081114, 17:44  #5  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:


20081114, 19:25  #6 
Oct 2007
100010_{2} Posts 
I agree this is essentially the same proof as Euclid's, this is reinventing the wheel, but you know what they great say "great minds think alike".

20081114, 20:01  #7 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
Did you know of Euclid's proof before you came up with this one? If you knew of it, I don't think that saying really applies.

20081114, 20:18  #8 
Oct 2007
2·17 Posts 
I've heard Euclid's proof but I didn't know exactly what it was,I came up
with this proof totally independantly his. 
20081114, 20:42  #9  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Quote:
(to get Euclid's, substitute "2*3*" for "3*", 1 for q, and X for 2^n, then argue about factors of X not among [2, ..., b]), but I like your formulation with the 2^n as an alternative. Last fiddled with by cheesehead on 20081114 at 20:48 

20081223, 15:30  #10 
Apr 2007
Spessart/Germany
A2_{16} Posts 
I think you still have to proof that there are at least an infinit number of solutions with q<>1.
Best regards Matthias 
20081223, 18:42  #11 
Aug 2002
Buenos Aires, Argentina
10100101101_{2} Posts 
That is not very difficult.
If for all members of the sequence q is equal to 1 we have: 3*5*7*11*...*p_{k} + 1 = u_{k} + 1 = 2^n for all k. u_{k} must be 3 (mod 4) for all k, but when p_{k+1} = 3 (mod 4) we get that u_{k+1} must be 1 (mod 4), so the next member of the sequence q cannot be equal to 1 in order to have a power of 2 at the right hand side. Last fiddled with by alpertron on 20081223 at 18:46 
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