I think I may really have got it now.<br><br>Since we can represent every situation like this by a doublesieve, we need only prove all doublesieves will have holes in them. Here is how I do this. A sieve of length n--let's say 16--has a set of spaces A that are at 2k, 3k, 5k...mk where k>1 and mk<n. It also has a set of spaces in the opposite manner, which can be given by nmod2+2k, nmod3+3k...nmodm+mk where once again k>1 and mk<n. All we need to prove is one necessary hole between those spaces. It is given thus: Choose a prime in the set that is under any 2k (let's choose 5) that has the property n mod p =1. We cannot create 5 from our 2k, 3k,spaces, because the only one to do so would be 5k, and k must be greater than 2. So we must find a nmodp+pk that is equal to 5, k of course being greater than 2 again. But alas! We find that the only way to have nmodp+pk=5 be if p was less than 5. so we have nmod2+2k which yields 4 and nmod3+3k which yields a minimum of seven! The point is here: Construct the sieve for any length. Choose a low prime. As long as none of the set B spaces are equal to it, we have a hole. The nmodp+pk must be equal to our P for the whole to be covered. But there is always a P inexpressible by nmodp+pk--simply choose a low p with a modulus of 1 for n. then, if nmodp is not 1, nmodp+pk is greater than P. But if nmodp IS one, nmodp+pk can never equal our P.<br>
<br><br><div class="gmail_quote">On Fri, Nov 27, 2009 at 9:30 PM, Will Steinberg <span dir="ltr"><<a href="mailto:asyluman@gmail.com" target="_blank">asyluman@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
I think I may have ended up merely restating something, but I think approaching the problem in a new paradigm is important--namely, instead of having each even number expressed as the sum of two primes, have each number n have 2 distinct primes an equal distance from it (because if Px + Py = 2n, then n - m = Px and n + m = Py. So I wrote out the numbers from a to a number 2n-1 (say 9) then wrote the numbers 9 to 1 below them, so each column sums to 2n (10).<br>
<br>e.g. 1 2 3 4 5 6 7 8 9 <br> 9 8 7 6 5 4 3 2 1<br><br> Then I ran a sieve of eratosthenes each way. The "holes" after a forwards and backwards run-through reveal the sets of primes which are equidistant from n and thus sum to 2n, or every even number. <br>
<br>Now, if we know there is one prime left unscathed by our doublesieve, it must have a mirror partner (also prime because of our sieving). And we KNOW that there is at least one prime p between n and 2n given Bertrand's postulate. Since this is a mirror, we can consider only one half of the line, from 1 to n. Now, we have already marked off the composites given by the sieve. But now lots are recast and even some primes must be lost. These are determined by taking whatever composite numbers are between n and 2n and subtracting them from 2n. Now we have a set that looks like this: {composites between 1 and n, 2n-composites between n and 2n}. <br>
<br>Visually, what we will have is a symmetrical pattern: two sieves laid on top of each other oppositeways. Each hole will have its conjugate and so the same for each filled in space. In essence, if we can prove that that pattern will always contain holes, then there will always be 2 primes able to add up to 2n. <br>
<br>I think the answer may lie with some sort of modulus thing. If you look at the center n, adding up prime numbers (to produce composites) will often lay a stretch of number over n. For example. the stretch of 9-12 lies over 10, with 1 on the right and 2 on the left. Consequently, all subsequent additions of 3 will give 10+2+3k. This means that the spaces with numbers that equal 10-2-3k will be filled in on the right half, as well as 10-4-7k, et cetera. I think the actual answer lies near and I would love for someone to give me insight (or tell me this has all already been done)<div>
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<br><br><br><div class="gmail_quote">On Fri, Nov 27, 2009 at 6:59 PM, spike <span dir="ltr"><<a href="mailto:spike66@att.net" target="_blank">spike66@att.net</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;"><br>
On Behalf Of Will Steinberg<br>
Subject: [ExI] Goldbach Conjecture<br>
<div><div></div><div><br>
<br>
I think I found something really good about the Goldbach<br>
Conjecture; does anyone have a background in number theory or know somebody<br>
who does? It is important<br>
<br>
-Will<br>
<br>
<br>
</div></div>Hi Will! I am a Goldbach conjecture fan. I know there are other<br>
Goldbachers here, Lee Corbin is one of our math superbrains, and several<br>
others too. What did you find?<br>
<font color="#888888"><br>
spike<br>
</font><div><div></div><div><br>
<br>
<br>
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