QuoteOriginally posted by: outrunAlso, the minimum set has nothing to do with Goldbach. I don't see the link.The challenge is to find products which equal E-p for every p using some subset of p. If you can do it with a subset, then p is not the minimum set. If p is the minimum set, then you can't do it and every even number is the sum of two primes.

Not sure how this got started but I think Goldbach aas been verified up to around a quintillion. Are there interesting wavelengths that long?

- Traden4Alpha
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QuoteOriginally posted by: farmerQuoteOriginally posted by: outrunAlso, the minimum set has nothing to do with Goldbach. I don't see the link.The challenge is to find products which equal E-p for every p using some subset of p. If you can do it with a subset, then p is not the minimum set. If p is the minimum set, then you can't do it and every even number is the sum of two primes.Interesting logic but I think the tricky bit is that both a subset of p might produce ALL the products needed for a given E, but still be the minimum set across all E. There are lots of small primes that can multiply to produce n and these can multiply in countless ways. The smallest n that uses 10 primes is only 59,049 and the smallest n that uses 20 primes is only 3,486,784,401 so you an see that there's a huge number of possible ways to make values of modest values of n.Perhaps I don't follow why "If you can do it with a subset, then p is not the minimum set" and why "if p is the minimum set, then you can't do it". It's not the set of p that forms n, its the powerset of p and that's a very big set.

Last edited by Traden4Alpha on September 16th, 2011, 10:00 pm, edited 1 time in total.

QuoteOriginally posted by: farmerQuoteOriginally posted by: outrunAlso, the minimum set has nothing to do with Goldbach. I don't see the link.The challenge is to find products which equal E-p for every p using some subset of p. If you can do it with a subset, then p is not the minimum set. If p is the minimum set, then you can't do it and every even number is the sum of two primes.Please, no offense, but you are wasting your time trying to come up with a clever simple proof. This is a very hard, deep problem, studied by the best mathematicians of the twentieth century.

QuoteOriginally posted by: exCBOEPlease, no offense, but you are wasting your time trying to come up with a clever simple proof. This is a very hard, deep problem, studied by the best mathematicians of the twentieth century.You are an idiot.1) You call it a "deep" problem. What does that mean? It means nothing. It is just an indication you have no idea what you are talking about.2) You say it was studied by the best mathematicians. You have no evidence of the actual time spent on it by each mathematician, nor any way to measure the relative qualifications of those who spent time. All you know is that some mathematicians studied it, but whatever they tried didn't work.3) I never tried to come up with a "simple proof." But nevertheless to me it is obvious the information properties of the series. As to how you would prove it using number theory, you probably cannot.4) I also think your suggestion that a "clever" proof is wrong, implying that hard work is necessary, is idiotic. People have put in hard work and failed. Obviously they are digging in the wrong spot. Instead of spending their lives digging, they should step back and look for a different place to dig, a clever place.5) How much time have I spent on it? If anybody wasted his time, it was these idiot mathematicians.

Last edited by farmer on September 17th, 2011, 10:00 pm, edited 1 time in total.

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