Goldbach's Conjecture - Proof (better link)

In summary: 2731 * 1708516002108856728977 = 3 * 7 * 19 * 31 * 103 * 1613 * 5689 * 2666396002108856728979 = 29 * 47 * 67 * 191 * 5119 * 872116002108856728981 = 11 * 17 * 193 * 241 * 383 * 5477 * 13029316002108856728983 = 3 * 83 * 389 * 1301 * 1590847585896002108856728985 = 5 * 1907 * 6439 * 141127 * 198
  • #1
Andy Lee
37
0
Hi. I might just have something here. Please let me know if you have any questions or comments. Thanks. Andy Lee. leeaj@shaw.ca

See proof at

http://www.facebook.com/pages/Goldbachs-Conjecture-Proof/237054761999
 
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  • #2
Can you upload better images so I can print it out to read?
 
  • #3
Better yet, make it LaTeX and upload it to arxiv.
 
  • #4
Unsurprisingly, the proof is flawed. On the last page, you make the assumption that the probabilities are 1/3, 1/3, and 2/3, respectively. First, this assumption is unjustified. Second, depending on how you define 'probability' in this context, it's probably false. Third, this assumption is not sufficient to prove the conjecture, even were it true.
 
  • #5
I have the last three pages on facebook now... I originally had duplicates of the first three.
 
  • #6
Ah, I see where you're going.

This approach is flawed. It is true that, for any bound B, a positive fraction of integers are relatively prime to all integers 2, 3, ..., B. It is even true that, for a given N, a positive fraction of integers are such that both n and N - n are relatively prime to all numbers 2, 3, ..., sqrt(N). But you need more than that: in particular, you need to show that n < N/2, or else the second term will be negative.

Let me put this another way, considering the simpler problem of finding primes in a range. Asymptotically, you expect at least 1/2 of numbers to be good mod 2, at least 2/3 of numbers to be good mod 3, at least 4/5 to be good mod 5, and at least 6/7 to be good mod 7. 1/2 * 2/3 * 4/5 * 6/7 = 8/35. So you might expect that out of
200, 201, 202, 203, 204, 205, 206, 207, 208
about 9 * 8/35 = 2.05... would be prime to 2, 3, 5, and 7. But in fact none are. This is because you can't guarantee that (out of N consecutive integers) N/2 numbers are odd; rather, you show that at least (N-1)/2 are. Likewise, not N/3 but at least (N-3)/3 are prime to 6, and so on. You haven't kept track of these additive portions, and you must!

In fact, if you properly track them, you'll see that as the number of primes you consider increases, your ranges quickly shrink to nothing. The approach doesn't work. I encourage you to work this out in detail so you can see this for yourself!
 
  • #7
Here's an example. Suppose S is a finite set of primes. Then, asymptotically,
[tex]k=\prod_{s\in S}1-\frac1s[/tex]
of numbers are relatively prime to all the primes in S ("candidates"). To get a nonasymptotic result, it's easy to see that every
[tex]P=\prod_{s\in S}s[/tex]
consecutive numbers give Pk candidates. Thus, at least
[tex]\lceil k(N-P-1)\rceil[/tex]
out of N consecutive numbers are candidates. Now, let S = {2, 3, 5, ..., sqrt(N)} so that the candidates, if less than N, are prime. Then
[tex]k=\prod_{p\le\sqrt N}1-\frac1p\approx\frac{2}{e^\gamma\log N}[/tex]
[tex]P=\prod_{p\le\sqrt N}p\approx\exp\left(\sqrt N\right)[/tex]
and so you have "at least" approximately
[tex]2\frac{N-\exp\left(\sqrt N\right)}{e^\gamma\log N}[/tex]
primes below N. But this is clearly negative!

So there's no obvious way to use this method to prove that there are any primes up to N/2, let alone primes that satisfy additional conditions.

You may object that the square root in the equation above could be replaced by something like a fourth root if the Riemann hypothesis holds. But (1) the RH hasn't been proven, and (2) the term is still far too large to work. (It looks like the Riemann hypothesis might be able to prove that at least 100-200 primes exist... but not even infinitely many primes, a truly ancient result!)
 
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  • #8
Thanks for the response. Concerning your example for the integers 201, 202, ... ,208:
I don't think it is necessarily relevant. My proof only requires that for any three consecutive odd integers, at least one will not be divisible by 3; for any 5 consecutive odd integers, at least one will not be divisible by 3 and 5; for any n consecutive odd integers, at least one will not be divisible by 3, 5, ..., n.

I think this is a perfectly valid statement.

Andy Lee
 
  • #9
Andy Lee said:
Thanks for the response. Concerning your example for the integers 201, 202, ... ,208:
I don't think it is necessarily relevant. My proof only requires that for any three consecutive odd integers, at least one will not be divisible by 3; for any 5 consecutive odd integers, at least one will not be divisible by 3 and 5; for any n consecutive odd integers, at least one will not be divisible by 3, 5, ..., n.

I think this is a perfectly valid statement.

1. You didn't prove this, so even if it was true your proof would fail.
2. It's false. Here are 47 consecutive odd numbers, each divisible by one of 3, ..., 47:
Code:
6002108856728919 = 3^3 * 11 * 17 * 1188771807631
6002108856728921 = 31 * 2711 * 71418817681
6002108856728923 = 29 * 1301 * 159084758587
6002108856728925 = 3 * 5^2 * 80028118089719
6002108856728927 = 37 * 181^2 * 4951593611
6002108856728929 = 7 * 857444122389847
6002108856728931 = 3 * 2000702952242977
6002108856728933 = 13 * 461700681286841
6002108856728935 = 5 * 647 * 1855365952621
6002108856728937 = 3^2 * 9719 * 50441 * 1360367
6002108856728939 = 43 * 870097 * 160423409
6002108856728941 = 11 * 71 * 127 * 9721 * 6224983
6002108856728943 = 3 * 7 * 1009 * 2003 * 8293 * 17053
6002108856728945 = 5 * 89 * 13487885071301
6002108856728947 = 23 * 701 * 372269978089
6002108856728949 = 3 * 83 * 24104854846301
6002108856728951 = 19 * 61 * 1861 * 2782749149
6002108856728953 = 17 * 101 * 5441 * 642472949
6002108856728955 = 3^2 * 5 * 74717 * 1785138547
6002108856728957 = 7 * 109 * 7907 * 8761 * 113557
6002108856728959 = 13 * 461700681286843
6002108856728961 = 3 * 97 * 199 * 367 * 282417587
6002108856728963 = 11 * 317 * 3217 * 535057997
6002108856728965 = 5 * 1117 * 3209 * 6277 * 53353
6002108856728967 = 3 * 2000702952242989
6002108856728969 = 41 * 1171 * 125015285179
6002108856728971 = 7 * 7927 * 36721 * 2945659
6002108856728973 = 3^4 * 1153 * 64267224061
6002108856728975 = 5^2 * 467 * 91939 * 5591743
6002108856728977 = 47 * 139 * 23627 * 38885047
6002108856728979 = 3 * 67 * 3457 * 8637905147
6002108856728981 = 29 * 100673 * 2055856793
6002108856728983 = 31 * 157 * 1233225571549
6002108856728985 = 3 * 5 * 7^2 * 11 * 13^3 * 919 * 367687
6002108856728987 = 17 * 727 * 436061 * 1113713
6002108856728989 = 19^2 * 653 * 25461470633
6002108856728991 = 3^2 * 666900984080999
6002108856728993 = 23 * 59 * 19069 * 231950921
6002108856728995 = 5 * 53 * 1709 * 5813 * 2279899
6002108856728997 = 3 * 26357 * 75907840507
6002108856728999 = 7 * 2144509 * 399832373
6002108856729001 = 37 * 151 * 797 * 3673 * 366983
6002108856729003 = 3 * 69593 * 28748623457
6002108856729005 = 5 * 1873 * 73303 * 8743279
6002108856729007 = 11 * 251 * 2173889480887
6002108856729009 = 3^2 * 11261 * 17551 * 3374291
6002108856729011 = 13 * 149939 * 3079256773
I can go higher if you like.
 
  • #10
The 47 consectutive numbers would have to appear before 47^2 for this to be a counter-example.

Andy Lee
 
  • #11
Andy Lee said:
The 47 consectutive numbers would have to appear before 47^2 for this to be a counter-example.

No, they wouldn't; that wasn't a part of your claim.

Why don't you write up precisely what you mean, prove it, then use that proof in your full proof of GC.
 
  • #12
You are correct, it was not part of my claim in the earlier post. But it most certainly is evident in the proof.

Andy Lee
 
  • #13
Andy Lee said:
You are correct, it was not part of my claim in the earlier post. But it most certainly is evident in the proof.

Your proof doesn't even state the required condition, let alone prove it. You need more specifics and less handwaving.

(Sorry, this comes off as harsh. But you need to do it for a valid proof!)
 
  • #14
Hey, CRGreathouse, can you translate his proof into English rather than picture? I have no clue what A1, A2,... B1, B2,... C1, C2,... are, and I believe it would easier to discuss the argument, and possibly any flaws, if it was written like this.

Cheers,

Rofler
 
  • #15
rofler said:
Hey, CRGreathouse, can you translate his proof into English rather than picture? I have no clue what A1, A2,... B1, B2,... C1, C2,... are, and I believe it would easier to discuss the argument, and possibly any flaws, if it was written like this.

If someone translated his whole argument into valid, gapless mathematics then the flaws would be obvious. But it's not easy to do that translation!
 
  • #16
CRGreathouse said:
If someone translated his whole argument into valid, gapless mathematics then the flaws would be obvious. But it's not easy to do that translation!

Wow, that's a pretty hostile comment! It also does not make a lot of sense, when you think about it. After one week and I have one person on this board who states the proof is flawed. Any other comments?
 
  • #17
Well, could you explain what A1, A2, B1, B2, etc. are? I can't comment on the quality of the proof if I don't understand/can't read/ don't want to turn my head 90 degrees to read the second page.

Cheers,

Rofler

By the way, saying that there is no set of n consecutive numbers UNDER n^2 each containing an odd factor less than n is quite a strong statement, and may even be stronger than the Goldbach conjecture itself. Of course the numbers n!+k each contain k as a factor, but n^2 is significantly stronger than n!. Or it may just be wrong, but I won't comment any further until I am able to understand the proof.

And CRGreathouse, despite apparently having valid arguments, I can't support either one of you until I see the original proof. You don't need to be so annoyed by this, at least he tried something, and apparently asides from omitting his endpoints, the proof from what you described would have been more or less correct.

Of course, the last sentence was more towards CRGreathouse's attitudes, I am always suspicious at purported "easy" proofs.
 
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  • #18
Andy Lee said:
Wow, that's a pretty hostile comment! It also does not make a lot of sense, when you think about it.

I spent the time to look over your proof and found numerous mistakes and gaps. When I finally understood your method and saw that the method cannot be used to prove the conjecture, I stopped looking further.

Most real mathematicians would refuse to even look at your proof -- there are lots of crackpots out there, and you would need to do something to demonstrate that you're not to convince them to spend their valuable time looking over the proof. Even going as far as I did was a gesture of good faith.

Andy Lee said:
After one week and I have one person on this board who states the proof is flawed. Any other comments?

Perhaps if you clean up the proof you can get rofler to look it over. He's clearly not a mathematician, but he may have useful input to give you.
 
  • #19
rofler said:
Well, could you explain what A1, A2, B1, B2, etc. are? I can't comment on the quality of the proof if I don't understand/can't read/ don't want to turn my head 90 degrees to read the second page.

If you're not willing to turn your head 90 degrees to read the purported proof, I think you'll understand if I'm not willing to attempt to translate it for you. :)

rofler said:
By the way, saying that there is no set of n consecutive numbers UNDER n^2 each containing an odd factor less than n is quite a strong statement, and may even be stronger than the Goldbach conjecture itself. Of course the numbers n!+k each contain k as a factor, but n^2 is significantly stronger than n!. Or it may just be wrong, but I won't comment any further until I am able to understand the proof.

I also don't see an easy way to show this. I'm not even sure if it's true, though I would suspect so.

rofler said:
And CRGreathouse, despite apparently having valid arguments, I can't support either one of you until I see the original proof. You don't need to be so annoyed by this, at least he tried something, and apparently asides from omitting his endpoints, the proof from what you described would have been more or less correct.

I'm not in the business of convincing you. I'd rather let you make your own decision, frankly. I looked the purported proof over and came to my own conclusion, and I chose to share that with this forum in case others were curious.

You may decide not to read it, in which case there's really not a whole lot to say.

You may read it and not be able to follow it; this really wouldn't be surprising given how it's written.

You may read it and conclude that it's wrong. In that case we agree and, again, there's not much to say.

*If* you chance to read it and, flabbergastingly decide that it's correct, perhaps I'll be interested enough to go back to the proof and point out the flaws in more detail. But I think I was explicit enough in post #4. Was that post not understandable, or do you not see why it applies?
 
  • #20
rofler said:
You don't need to be so annoyed by this, at least he tried something, and apparently asides from omitting his endpoints, the proof from what you described would have been more or less correct.

I did want to comment on this, at least. It is ABSOLUTELY not true. This may seem like a small point in the proof, but it actually causes the proof to fail. In particular, the error term becomes larger than the main term, so instead of proving that there are at least f * N Goldbach partitions, where f tends to 0 but f*N increases without bound, it shows that at least f * N - O(stuff) Goldbach partitions, where O(stuff) actually becomes larger than f * N (and indeed, larger even than N).

It is a fatal flaw, not an accounting error.
 
  • #21
You're missing the point... I know there is a flaw, and it is quite obvious where it is based on what you said... I just wanted to understand what he was TRYING... he's clearly not a mathematician, however I am certainly qualified enough in mathematics to understand every word that you said, so don't give me any of that.

Furthermore, not that it matters, but if you remember chinese remainder theorem, you could realize that just adding n! to each of the first n consecutive numbers yields a sequence of numbers so that each is divisible by at least one of 3,5,...,n.

I just wanted to participate in the discussion, and in order to do that, I just wanted to know what HE was saying. Of course, as I said before, I believe 100% that you found a flaw in the proof, I just wanted to see it for myself.

Also, after looking at your comments completely, I understand enough about where the flaws are to get the jist of the proof.

Cheers,

Rofler
 
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  • #22
rofler said:
Furthermore, not that it matters, but if you remember chinese remainder theorem, you could realize that just adding n! to each of the first n consecutive numbers yields a sequence of numbers so that each is divisible by at least one of 3,5,...,n.

I don't see how Sun-Zi's theorem (the chinese remainder theorem, as you say) is relevant, but I certainly understand what you're getting at. (What you say is not literally true: n! + 1, n! + 2, n! + 4, etc. are not divisible by any of 3, 5, ..., n.) But as you've already pointed out, this gives a far weaker result than that claimed by the OP.

rofler said:
I just wanted to participate in the discussion, and in order to do that, I just wanted to know what HE was saying. Of course, as I said before, I believe 100% that you found a flaw in the proof, I just wanted to see it for myself.

I would love for the OP to explain himself. It would be nice if he understood the flaw in the proof, because then he could work toward a valid proof. Not that I expect him to find it -- the chance that any given person will discover the proof is of course quite low -- but that would be a step in his mathematical journey, and he may well come to prove other interesting things. (And of course if after understanding his mistake he *did* manage to correctly prove the conjecture, all the better!)

rofler said:
Also, after looking at your comments completely, I understand enough about where the flaws are to get the jist of the proof.

Glad to have helped. I don't suppose you have a good way to explain what I wrote to the OP? Communication is hard.
 
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  • #23
Ah, I meant n!+3 n!+5, ..., n!+n, but in retrospect, it was meant n consecutive numbers, not n consecutive odd numbers, so chinese remainder would not quite work, sorry (I was thinking of setting up the congruences a=0 mod 3, a+2=0 mod 5, a+4=0 mod 7,... but that was before I realized it was n consecutive numbers)

By the way, not that it is relevant, but this teacher goes to my school. Hey Mr. Lee.

Anyways, I think what you said is very clear, even without the knowledge of asymptotes, if the OP decided to redo his proof with considering the endpoints, he would either get a lower bound that does not stay positive, or get stuck on proving that in any consecutive string of n numbers below n^2, at least one is relatively prime to every odd number less than n.

Cheers,

Rofler
 
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  • #24
rofler said:
Anyways, I think what you said is very clear, even without the knowledge of asymptotes, if the OP decided to redo his proof with considering the endpoints, he would either get a lower bound that does not stay positive, or get stuck on proving that in any consecutive string of (n-1)/2 numbers below n^2, at least one is relatively prime to every odd number less than n.

If the proof was redone considering the error terms, it could be used to prove that for every set S of primes, there exists some N(S) such that for all n > N, there exists an m such that m and n - m are relatively prime to all primes in S. If N({2, 3, 5, ..., p}) was smaller than p^2, this would prove Goldbach's conjecture. But the OP has not considered the error term that would allow N to be computed. As I posted some time ago, the naive error term is exponential in p and the error term using the (unproven) RH is exponential in sqrt(p). These are far, far larger than what would be required to prove GC.
 
  • #25
But what he then proved if it was corrected is really quite simple, isn't it? Take N(S)= lcm(all primes in S), and then if p_1,p_2,...,p_k are the primes in this set that divide n, take m to be 1 mod each of these primes and either 1 or 2 mod the rest, so that m and n-m are relatively prime to the prime in question (by CRT, and the condition that n>lcm(all primes in S), this is possible). Too bad the bound didn't work. Note that it is obviously incorrect if 2 is in the set of primes, just take n odd. Ah well, too bad.

Cheers,

Rofler
 
  • #26
Okay, here is an example of what I was trying to generalize in my proof.

Let’s say N=37
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73
becomes
x, 5, 7, x, 11, 13, x, 17, 19, x, 23, 25, x, 29, 31, x, 35, 37, x, 41, 43, x, 47, 49, x, 53, 55, x, 59, 61, x, 65, 67, x, 71, 73
when we remove multiples of 3, and
x, y, 7, x, y, 13, x, y, 19, x, y, 25, x, y, 31, x, y, 37, x, y, 43, x, y, 49, x, y, 55, x, y, 61, x, y, 67, x, y, 73
when we remove pairings with x that sum to 74 (2*37).
Then 1/3 of the numbers remain. 1/3 is the minimum that can remain.

Do the same with multiples of 5.
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73
becomes (I’ll use ‘a’ and ‘b’ in place of x and y)
3, a, 7, b, 11, 13, a, 17, b, 21, 23, a, 27, b, 31, 33, a, 37, b, 41, 43, a, 47, b, 51, 53, a, 57, b, 61, 63, a, 67, b, 71, 73
Then 3/5 of the numbers remain. 3/5 is the minimum that can remain.

Do the same with multiples of 7.
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73
becomes (I’ll use ‘c’ and ‘d’ in place of x and y)
3, 5, c, 9, d, 13, 15, 17, 19, c, 23, d, 27, 29, 31, 33, c, 37, d, 41, 43, 45, 47, c, 51, d, 55, 57, 59, 61, c, 65, d, 69, 71, 73
Then 5/7 of the numbers remain. 5/7 is the minimum that can remain

Put them all together…
x, ay, c, bx, dy, 13, ax, y, b, cx, y, ad, x, by, 31, x, acy, 37, bdx, y, 43, ax, y, bc, x, dy, a, x, by, 61, cx, ay, d, bx, y, 73
Then at least 1/3*3/5*5/7 = 1/7 of the numbers will remain.

18 possible pairs. 1/7*18 >1. There is at least one prime pairing that sums to 74.
In fact, there are three… 13+61, 31+43, 37+37

This is a specific example illustrating the general proof that I have shown on facebook.
Can you show me a counter-example demonstrating how this fails?
 
  • #27
For reference, your first mistake is here:
Andy Lee said:
Then 1/3 of the numbers remain. 1/3 is the minimum that can remain.

In fact, out of N consecutive odd numbers, as few as (N - 2)/3 might remain.

Andy Lee said:
Can you show me a counter-example demonstrating how this fails?

Some small counterexamples are
N = 5, p = 3
N = 9, p = 5
N = 39, p = 7
 
  • #28
It may also be instructive to note that you don't use the fact that the sequence starts at 3. Therefore, were the proof valid, you could choose to start at any higher number and the proof would go through. Thus the method would expect at least 3 numbers from my post #9 'counterexample' to not only be relatively prime to 3, 5, 7, ..., 47, but also to have 2N minus those numbers be relatively prime to 3, 5, ..., 47 as well.

(For reference, my example was the odds from 6002108856728919 to 6002108856729011 inclusive; see the first page for factorizations.)

I say this to help you understand this aspect of the proof better. If you can understand how it is that these numbers fail (by all having small divisors), you should be able to see the gap in your proof. And once you see it, you may be able to fix it!

Obviously demonstrating by examples that a supposed proof of GC is wrong is difficult, because we don't believe that a counterexample to GC exists. So I focus entirely on counterexamples to the method rather than GC.
 
  • #29
Thanks. I now see what you mean by O(stuff) in your earlier post.
 
  • #30
Andy Lee said:
Thanks. I now see what you mean by O(stuff) in your earlier post.

I may not have explained it well before. I tried...
 
  • #31
CRGreathouse said:
For reference, your first mistake is here:


In fact, out of N consecutive odd numbers, as few as (N - 2)/3 might remain.


For each iteration we could say as few as (N-M)/D might remain, where multiples of
D are eliminated and M is the maximum divisor in the process.

Then near the end of my proof instead of the expected number being
(1/sqrt(2N))*((N-1)/2)) I would have

(1/sqrt(2N))*((N-M)/2)) and M<=sqrt(2N) so N-M >= N-sqrt(2N) giving
an expected number of prime pairs greater than or equal to

(1/sqrt(2N))*((N-sqrt(2N)/2)

and this is >1 for N>=18.

So the proof is good for N>=18
 
  • #32
Andy Lee said:
M<=sqrt(2N)

Care to explain the derivation of that?
 
  • #33
My previous counterexample with
6002108856728919
still applies. (47 - sqrt(2 * 47)) * 1/3 * 3/5 * ... * 45/47 > 1, so you've 'proved' that that interval contains at least 2 odd numbers with smallest prime factor > 47.
 
  • #34
CRGreathouse said:
My previous counterexample with
6002108856728919
still applies. (47 - sqrt(2 * 47)) * 1/3 * 3/5 * ... * 45/47 > 1, so you've 'proved' that that interval contains at least 2 odd numbers with smallest prime factor > 47.

Again, as I stated before, the counterexample is not relevant unless it was less than 47^2
 
  • #35
Then you must prove that he cannot create a counterexample.

So first and foremost, if you are using the fact that his counterexample does not work because it is larger than n^2, then you must then prove the following statement for your proof to be valid:

We cannot have n consecutive numbers between 1 and n^2 so that each is divisible by at least one of 3,5,...,n.

After that, the other points of the proof can be discussed.

Cheers,

Rofler
 

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