mersenneforum.org "Variations on a theme by Euclid"
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 2008-11-14, 15:18 #1 Carl Fischbach     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.
 2008-11-14, 15:55 #2 10metreh     Nov 2008 2×33×43 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 no-one has thought of it before? I've never seen it before. Last fiddled with by 10metreh on 2008-11-14 at 15:56
 2008-11-14, 16:56 #3 Carl Fischbach     Oct 2007 428 Posts This definitely a new proof unless I accidently reproduced somsone elses work I've never seen before.
 2008-11-14, 17:07 #4 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 10000101010112 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 Mini-Geek on 2008-11-14 at 17:47 Reason: Euler, Euclid, what's the difference? both old guys with "Eu" in their names...;-)
2008-11-14, 17:44   #5
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by Mini-Geek I'm not sure, but it sounds to me like a roundabout version of Euler's proof of infinite primes. http://en.wikipedia.org/wiki/Prime_n..._prime_numbers
I think you mean Euclid

 2008-11-14, 19:25 #6 Carl Fischbach     Oct 2007 1000102 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".
2008-11-14, 20:01   #7
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17×251 Posts

Quote:
 Originally Posted by Carl Fischbach 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".
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.

 2008-11-14, 20:18 #8 Carl Fischbach     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.
2008-11-14, 20:42   #9

"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·641 Posts

Quote:
 Originally Posted by Carl Fischbach 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.
It's very closely related to Euclid's proof

(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 2008-11-14 at 20:48

 2008-12-23, 15:30 #10 MatWur-S530113     Apr 2007 Spessart/Germany A216 Posts I think you still have to proof that there are at least an infinit number of solutions with q<>1. Best regards Matthias
 2008-12-23, 18:42 #11 alpertron     Aug 2002 Buenos Aires, Argentina 101001011012 Posts That is not very difficult. If for all members of the sequence q is equal to 1 we have: 3*5*7*11*...*pk + 1 = uk + 1 = 2^n for all k. uk must be 3 (mod 4) for all k, but when pk+1 = 3 (mod 4) we get that uk+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 2008-12-23 at 18:46

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