EFF claim(s)
 

Please indulge me with this.
I don't necessarily require comment, and I'm not being deliberately confrontational - I just want to put down a marker (I wrote directly to the EFF about this last year, and never received a reply). I want that prize money(s)!!! :-)
The topic, however contentious, is at least relevant to this forum...
[url]http://www.bearnol.pwp.blueyonder.co.uk/Math/mmprime.html[/url]
Thanks,
J

Step 2 of this proof appears to fall down as it assumes there are an infinite number of mersenne primes.

There ARE an infinite supply of mersenne NUMBERS. Most of these are NOT prime.

There does not yet exist a proof that the supply of mersenne PRIMES is infinite.

They will either need to SHOW this to be true, or rethink the supposed proof.

Regards,

Peter

Hi Peter,
Thanks for your response.
The link quoted ([url]http://www.bearnol.pwp.blueyonder.co.uk/Math/perfect2.htm[/url]) [from within the link quoted above originally by me] yields a proof of the Inf. of Mersenne primes (ie it is a lemma on the way to my proof).
J

[QUOTE=bearnol]Hi Peter,
Thanks for your response.
The link quoted ([url]http://www.bearnol.pwp.blueyonder.co.uk/Math/perfect2.htm[/url]) [from within the link quoted above originally by me] yields a proof of the Inf. of Mersenne primes (ie it is a lemma on the way to my proof).
J[/QUOTE]Except that "proof" isn't. Here is the so-called proof:
[code]There are an infinite number of Mersenne primes (2**p-1) and hence (even) perfect numbers.

Proof:

Suppose Mp = 2**p-1

hcf (Mp, n) = 1, for all n<p [by Lemma 1]

let p->99999… [Euclid]

=> Mp prime

=> there exist an infinite number of prime Mp[/code]

What on earth does "let p->99999... [Euclid]" mean? All the previous line shows is that if Mp has a prime factor, it can not be less than p. You have not shown that it can not be less than Mp.

Paul

P.S. Sometimes I wonder why I bother. My only excuse is that I feel that I may be doing others a favour when I draw attention to sloppy reasoning in a purported proof, in the hope that they may pay attention and avoid such themselves.

Your claim "There are an infinite number of Mersenne primes (2**p-1) and hence (even) perfect numbers."

You appear to be presenting something as a proof that there are infinite mersenne primes.

I already believe your LEMMA1 "Any divisor of Mp is of the form 2kp+1"
And further you seem to use LEMMA2 only to prove this.

So now if I try to follow your reasoning, you use LEMMA1 (which I accept) to prove your assertion using the following...


"Proof:

Suppose Mp = 2**p-1

hcf (Mp, n) = 1, for all n<p [by Lemma 1]

let p->99999… [Euclid]

=> Mp prime

=> there exist an infinite number of prime Mp"


Your line "let p->99999...[Euclid]" appears to be a reference to the fact we know and have proof that prime numbers are infinite.

You appear to be proving that any mersenne prime is a prime number (which is self evident by definition), rather than that if we keep checking primes (or mersenne numbers) some of them will be mersenne primes.

Sorry if I misunderstand, my computing is superior to my maths.

But from my reading of it I don't think this is proof that there exist infinite number of prime Mp.

Paul, I'm with you.

Suggestion: If your thinking isn't yet an actual proof then call it something like "towards a proof" or you will be ridiculed by Dr Silverman ;-)

P.S. I did like your Java code in the other controversial thread, even if your algorithm wasn't faster. A language like C would probably be understood by a wider audience here though.

I could see why you didn't get a reply.

Hi Peter (and to a lesser extent(!) Paul),
(Thanks again for your response) [and also thanks for the positive comment re my Java, which I _do_ like as a language, though I know a little C too]
I'll try and keep responding as I see points I still want to argue, just folks let me know if /when I should stop...
As regards my linked work. I do genuinely claim (and believe) this is a complete proof [and the exposition is moreorless complete - if you look at other proofs on my site, listed above this one, I give slightly more detailed steps about this particular/similar technique at its first occurrence, which might aid understanding - also I'll expand below...]. However I grant this step you pick up on here is slightly contentious. I've had lengthy and detailed discussions about this proof with Will Edgington last year by email, and he failed to convince me I was wrong (though admittedly I suspect I failed to totally convince him I was right :-) - anyway I think we both learnt something from the exchange)
...speaking of which I _do_ (so far in my brief experience) like _this_ forum. I've had the best discussions (of any forum at least) yet re math, here... anyway I'll try not to waffle and dilute the/anyone's experience...
You surmise correctly that I use the infinitude of primes [Euclid].
What I'm saying is that since any divisor of Mp, as p->inf, has to be strictly greater than p [actually at least as big as 2kp+1], and since it is impossible to have a number _strictly_ greater than infinity, then there can be no proper divisors of Mp 'at infinity', but instead an infinite string of prime Mp.

This boggles the mind. I honestly don't know where to begin.

I suppose Bob must be wincing in pain reading these threads... :ermm:

Alex

I'm wincing and I'm not a mathematician.

The world has a lot of dogs.
Because the world has a lot of dogs, they can't possibly be counted.
The only representation of an uncountable number is infinity.
Therefore the number of dogs in the world is infinite.

Hey, its just as good as the logic above!

Fusion :bounce:

Can you prove that the number of dogs in this universe are finite? :devil:

[QUOTE=Citrix]Can you prove that the number of dogs in this universe are finite? :devil:[/QUOTE]I can’t prove that there are only a finite number of dogs in the universe, but I think I can prove that there are only a finite number of dogs on earth.

A dog is a mammal.
Mammals reproduce by giving birth to live young.
Therefore dogs reproduce by giving birth to live young.
A dog lives for a finite number of years.
It takes a finite period of time for a dog to reproduce in this manner, and therefore a dog can only produce a finite number of puppies in its lifetime.
Since at some point in time there was only one pair of dogs on earth, which produced only a finite number of puppies, which in turn produced only a finite number of puppies, which in turn produced only a finite number of puppies, there have only ever been born a finite number of puppies. Because dogs live for a finite number of years, some of these have died.
Therefore, there are a finite number of dogs on earth.

We do naturally have to accept that “dog” in this context includes bitches.