Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Our Old Friend, the Twin Primes Conjecture

  1. Oct 23, 2014 #1
    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.
     
  2. jcsd
  3. Oct 23, 2014 #2

    jedishrfu

    Staff: Mentor

    If you have a plausible proof why not discuss it with one of your math professors? Perhaps it could get posted on the Arxiv and get greater exposure to people who can really review it.
     
  4. Nov 17, 2014 #3
    I can propose the following criteria of twin primes conjecture:
    Natural numbers N1=6n+5 and N2=6n+7, n=0,1,2,3,..
    are twins if and only if no one of three equations
    n=6xy-x+y-1; x>=1; y>=1
    n=6xy-x-y-1; x>=1; y>=x;
    n=6xy+x+y-1; x>=1; y>=x;
    has integer solution.
    Attached: convenient C++ program for finding primes
     

    Attached Files:

  5. Nov 17, 2014 #4

    jedishrfu

    Staff: Mentor

    Is this a personal theory? or are you looking for help with your program?
     
  6. Nov 17, 2014 #5
    Yes, this criteria of twin primes based on derived "alternative definition" of prime numbers:
    Natural numbers that do not appear in arrays
    P1(i,j)=6*i*j-i+j-1; i=1,2,3,...; j>=i
    P2(i,j)=6*i*j+i-j-1; i=1,2,3,...; j>=i+1
    are indexes p of ALL PRIMES in the sequence S1(p)=6*p+5, p=0,1,2,3,...
    Natural numbers that do not appear in arrays
    P3(i,j)=6*i*j-i-j-1; i=1,2,3,...; j>=i
    P4(i,j)=6*i*j+i+j-1; i=1,2,3,...; j>=i
    are indexes p of ALL PRIMES in the sequence S2(p)=6*p+7, p=0,1,2,3,...
    Attached C++program was successfully tested for N up to 2*10^18
     
    Last edited: Nov 17, 2014
  7. Nov 17, 2014 #6

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    This may or may not give all twin primes (you haven't given a proof or a reference to one), but assuming it does, how does it show that there are an infinite number of twin primes?
     
  8. Nov 17, 2014 #7
    Example. n=37. N1=S1(37)=37*6+5=227; N2=S2(37)=37*6+7=229
    37=6*x*y-x+y-1; y=(38+x)/(6*x+1); x=1; y=39/7; x=2;y=40/13; x=3;y=41/19;....no integer solution
    37=6*x*y-x-y-1; y=(38+x)/(6*x-1); x=1; y=39/5; x=2;y=40/11; x=3;y=41/17;....no integer solution
    37=6*x*y+x+y-1; y=(38-x)/(6*x+1); x=1; y=37/7; x=2;y=36/13; x=3;y=35/19;....no integer solution
    N1, N2- twin primes
    n=36. N1=S1(36)=36*6+5=221; N2=S2(36)=36*6+7=223
    36=6*x*y-x+y-1; y=(37+x)/(6*x+1); x=1; y=38/7; x=2;y=39/13=3 - integer solution
    36=6*x*y-x-y-1; y=(37+x)/(6*x-1); x=1; y=37/5; x=2;y=38/11; x=3;y=39/17;....no integer solution
    36=6*x*y+x+y-1; y=(35-x)/(6*x+1); x=1; y=34/7; x=2;y=33/13; x=3;y=32/19;....no integer solution
    N1, N2- not twin primes
     
  9. Nov 17, 2014 #8

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yorshick, we're not looking for more examples. We're looking for (a) a proof that your rule applies to *all* twin primes, and (b) an explanation of how your rule shows that there are an infinite number of twin primes, which is the issue under discussion.
     
  10. Nov 17, 2014 #9

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Counterexample: There is no solution for N=122, but 6N+5=737 is not prime.

    The first claim in the first post also has a counterexample that is more complicated to construct.
    Edit: To extend that: choose an arbitrary value of Q. No matter what S we choose, D<Q so we have a finite number of possible D where Q-D is prime (no pair needed). Now all we have to do is to find an odd number S such that S-D divides all possible values of Q-D.
    As all values of Q-D are primes by construction, this is always possible and will give an counterexample.
     
    Last edited: Nov 17, 2014
  11. Nov 18, 2014 #10
    n=122. N1=S1(122)=6*122+5=737; N2=S2(122)=6*122+7=739

    122=6*x*y-x+y-1; y= (123+x)/(6*x+1); x=1; y=124/7; x=2;y=125/13; x=3;y=126/19; x=4;
    y=127/25; x=5;y=128/31.... x=11; y=134/67 = 2 - integer solution
    N1=737 and N2=739 are not twin primes
     
    Last edited: Nov 18, 2014
  12. Nov 18, 2014 #11
    Proof of criteria of twin primes is contained in attached file.
    It follows from primality criteria (section 4).
     

    Attached Files:

    Last edited: Nov 18, 2014
  13. Nov 18, 2014 #12

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Sorry, missed that number in the collection of solutions.
    Anyway, this is not the right place for original research. If this is published research please give a reference, then we can discuss it (in a new thread). Otherwise, it is not an acceptable reference until published.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Our Old Friend, the Twin Primes Conjecture
Loading...