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Math Help - A Proof & Goldbach Conjecture

  1. #1
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    A Proof & Goldbach Conjecture

    My prof. said this would be the hardest prob on my hw .

    1a.) For all integers n >= 1, prove there exists integers a and b such that tau(a) + tau(b) = n, where the number of divisors func., denoted by tau, is defined by setting tau(n) equal to the # of pos. divisors of n.

    b.) Now prove the Goldbach conjecture says that for each even integer 2n, there exists integers a and b such that sigma(a) + sigma(b) = 2n, where the sum of the divisors func., denoted by sigma, is defined by setting sigma(n) equal to the sum of all pos. divisors of n.

    (Note that tau/sigma are multiplicative functions).
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  2. #2
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    Quote Originally Posted by flash101 View Post
    My prof. said this would be the hardest prob on my hw .

    1a.) For all integers n >= 1, prove there exists integers a and b such that tau(a) + tau(b) = n, where the number of divisors func., denoted by tau, is defined by setting tau(n) equal to the # of pos. divisors of n.
    Here is the first one.
    Attached Thumbnails Attached Thumbnails A Proof & Goldbach Conjecture-picture18.gif  
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  3. #3
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    Quote Originally Posted by flash101 View Post

    b.) Now prove the Goldbach conjecture says that for each even integer 2n, there exists integers a and b such that sigma(a) + sigma(b) = 2n, where the sum of the divisors func., denoted by sigma, is defined by setting sigma(n) equal to the sum of all pos. divisors of n.
    This one is easy with Goldbach Conjecture.

    This example is trivial with 2n=2 just pick a=1 and b=1.

    Say 2n>2 meaning at least 4.

    Then, 2n-2 is an even positive integer above two.

    Then by Goldbach conjecture there shall exist primes p and q such as:
    p+q=2n-2

    Then choose p=a and q=b.
    Because,
    o(a)=o(p)=p+1
    o(b)=o(q)=q+1
    Thus,
    o(a)+o(b)=p+1+q+1=(p+q)+2=2n-2+2=2n.
    Q.E.D.
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  4. #4
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    Oddly, my professor was amazed at how you did part b. She had done it a totally different way (much harder), but said it was a very good way you did it. Part b is supposedly harder than part a, and I showed her your way but she said there was an easier way than that, but it would work. I, however, could not get q_1, g (I don't even know where g comes from!) in your proof and show tau(m) = n - 1 . I'm so horrible at proofs. You must do very well on proofs in your math classes. If you get time can you help me finish the proof?
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  5. #5
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    Quote Originally Posted by flash101 View Post
    Oddly, my professor was amazed at how you did part b.
    Thank you.
    She had done it a totally different way (much harder), but said it was a very good way you did it.
    Of course, she is a woman. They always complicate things .
    Part b is supposedly harder than part a, and I showed her your way but she said there was an easier way than that, but it would work.
    Did you say "ThePerfectHacker says that ...." in the middle of class?

    I, however, could not get q_1, g (I don't even know where g comes from!) in your proof and show tau(m) = n - 1 .
    Okay, my proof to #1 is really messy. Basically what I did was this. We want to show we can find a and b. So I simply say we can always choose a=1 and then find a corresponding "b" which works. The question is, of course, how do you find b. And I give a constriction. The meaning of p_l,g is just a notation meaning all the primes I listed, say, p_2,3 and so one. I just used the letters "g" and "l" is because there was so many letters flying around I had to use some strange ones.

    I'm so horrible at proofs.
    Bad Profs lead to bad proofs.
    You must do very well on proofs in your math classes.
    I am a fashion designer .
    If you get time can you help me finish the proof?
    Try doing this first. Use the "tao" formula for finding how many divisors a positive integer by using it on my long constructed number. It is already in factored form.
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