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Old 2008-11-14, 15:18   #1
Carl Fischbach
 
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Default "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.
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Old 2008-11-14, 15:55   #2
10metreh
 
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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
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Old 2008-11-14, 16:56   #3
Carl Fischbach
 
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This definitely a new proof unless I accidently reproduced somsone elses
work I've never seen before.
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Old 2008-11-14, 17:07   #4
Mini-Geek
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"Tim Sorbera"
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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...;-)
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Old 2008-11-14, 17:44   #5
davieddy
 
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Quote:
Originally Posted by Mini-Geek View Post
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
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Old 2008-11-14, 19:25   #6
Carl Fischbach
 
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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".
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Old 2008-11-14, 20:01   #7
Mini-Geek
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Quote:
Originally Posted by Carl Fischbach View Post
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.
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Old 2008-11-14, 20:18   #8
Carl Fischbach
 
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I've heard Euclid's proof but I didn't know exactly what it was,I came up
with this proof totally independantly his.
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Old 2008-11-14, 20:42   #9
cheesehead
 
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Quote:
Originally Posted by Carl Fischbach View Post
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
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Old 2008-12-23, 15:30   #10
MatWur-S530113
 
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I think you still have to proof that there are at least an infinit number of solutions with q<>1.

Best regards

Matthias
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Old 2008-12-23, 18:42   #11
alpertron
 
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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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