"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.

You have to make it clear that 2 is not included in the sequence of primes.

Otherwise, good proof! :showoff:

P.S. Are you sure no-one has thought of it before? I've never seen it before.

This definitely a new proof unless I accidently reproduced somsone elses
work I've never seen before.

I'm not sure, but it sounds to me like a roundabout version of Euclid's proof of infinite primes.
[URL]http://en.wikipedia.org/wiki/Prime_number#There_are_infinitely_many_prime_numbers[/URL]

[quote=Mini-Geek;149326]I'm not sure, but it sounds to me like a roundabout version of Euler's proof of infinite primes.
[URL]http://en.wikipedia.org/wiki/Prime_number#There_are_infinitely_many_prime_numbers[/URL][/quote]
I think you mean Euclid

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".

[quote=Carl Fischbach;149350]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".[/quote]
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.

I've heard Euclid's proof but I didn't know exactly what it was,I came up
with this proof totally independantly his.

[quote=Carl Fischbach;149305]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.[/quote]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. :smile:

I think you still have to proof that there are at least an infinit number of solutions with q<>1.

Best regards

Matthias

That is not very difficult.

If for all members of the sequence q is equal to 1 we have:

3*5*7*11*...*p[sub]k[/sub] + 1 = u[sub]k[/sub] + 1 = 2^n for all k.

u[sub]k[/sub] must be 3 (mod 4) for all k, but when p[sub]k+1[/sub] = 3 (mod 4) we get that u[sub]k+1[/sub] 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.