- #1
tuttlerice
- 28
- 0
I have to be honest--- I am not sure exactly the right tone to strike here. I find that if one comes in cocksure proclaiming "I have a proof of the twin primes conjecture! SOLVED! QED, BAY-BAY!" then one achieves a great deal of annoyance, and rightly so. On the other hand, it also seems like "well, I'm probably wrong, and I just want to learn from the experience of having my proof shot down" equally achieves annoyance.
I really do not want to annoy anyone. I swear. I just want to strike a posture that is somewhere in between--- I think I have something worth looking at, but I concede it could be wrong, because people make mistakes.
But I've written what I hope to be a proof that I think is promising. It does not involve sieves; it is not another variation on Euclid's multiply-them-all-together-and-add-1 approach. I've read pretty much every false proof on here and I do see how these repeatedly asserted mistakes can annoy.
The essential structure is as follows:
1) Consider the numbers S where S is an odd not divisible by any twin prime and Q that is a non-twin prime > 3.
2) Prove that given S and Q as described in (1) there exists a number D such that Q-D is a twin prime and S-D is not divisible by Q-D.
3) Prove that where S-Q=A, A is not divisible by some twin prime. This is because A=(S-D)-(Q-D), and (S-D)-(Q-D) is not divisible by one particular twin prime (namely, Q-D: Q-D is divisible by Q-D but S-D is not).
4) This means given S and Q as constructed, if A + Q = S, then A is not divisible by some twin prime.
5) Assume there are finitely many twin primes.
6) Consider P# as a primorial (including 2 as a factor) that includes all the finitely many twin primes as factors.
7) Consider P# + Q, where Q is a non-twin prime > 3. Its sum, S, is an odd not divisible by any twin prime because P# is divisible by all of them but Q is not divisible by any of them.
8) Observe that P# + Q = S follows the form A + Q = S as described in (4).
9) This means P# is not divisible by some twin prime, per (4).
10) This is a contradiction--- P# by definition is divisible by all the twin primes. The only resolution is that the supposition of (5) must be incorrect, and there are infinitely many twin primes.
I made a youtube video with the details (i.e., a formal proof is presented there).
My e-mail address and real name are in the video. While obviously I'm not against discussing it here if people are interested, I'd actually prefer discussing it with real people under real names privately if anyone cares to. I find people are much more considerate and compassionate when they are not hiding behind pseudonyms. (Yes, I'm a little insecure, I admit it.) Also, I promise not to argue tediously and relentlessly once someone points out a flaw to me. I sympathize with those who have been embroiled in those frustrating kinds of conversations.
Thank you for your time.
I really do not want to annoy anyone. I swear. I just want to strike a posture that is somewhere in between--- I think I have something worth looking at, but I concede it could be wrong, because people make mistakes.
But I've written what I hope to be a proof that I think is promising. It does not involve sieves; it is not another variation on Euclid's multiply-them-all-together-and-add-1 approach. I've read pretty much every false proof on here and I do see how these repeatedly asserted mistakes can annoy.
The essential structure is as follows:
1) Consider the numbers S where S is an odd not divisible by any twin prime and Q that is a non-twin prime > 3.
2) Prove that given S and Q as described in (1) there exists a number D such that Q-D is a twin prime and S-D is not divisible by Q-D.
3) Prove that where S-Q=A, A is not divisible by some twin prime. This is because A=(S-D)-(Q-D), and (S-D)-(Q-D) is not divisible by one particular twin prime (namely, Q-D: Q-D is divisible by Q-D but S-D is not).
4) This means given S and Q as constructed, if A + Q = S, then A is not divisible by some twin prime.
5) Assume there are finitely many twin primes.
6) Consider P# as a primorial (including 2 as a factor) that includes all the finitely many twin primes as factors.
7) Consider P# + Q, where Q is a non-twin prime > 3. Its sum, S, is an odd not divisible by any twin prime because P# is divisible by all of them but Q is not divisible by any of them.
8) Observe that P# + Q = S follows the form A + Q = S as described in (4).
9) This means P# is not divisible by some twin prime, per (4).
10) This is a contradiction--- P# by definition is divisible by all the twin primes. The only resolution is that the supposition of (5) must be incorrect, and there are infinitely many twin primes.
I made a youtube video with the details (i.e., a formal proof is presented there).
My e-mail address and real name are in the video. While obviously I'm not against discussing it here if people are interested, I'd actually prefer discussing it with real people under real names privately if anyone cares to. I find people are much more considerate and compassionate when they are not hiding behind pseudonyms. (Yes, I'm a little insecure, I admit it.) Also, I promise not to argue tediously and relentlessly once someone points out a flaw to me. I sympathize with those who have been embroiled in those frustrating kinds of conversations.
Thank you for your time.