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Math Help - Distribution of primes

  1. #1
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    Distribution of primes

    Show that for every positive integer N there is an even number K so that there are more than N pairs of successive primes such that K is the difference between these successive primes (Hint: Use prime number theorem)

    So, this is how far I've gotten..

    Consider a list of primes and consider the difference of the successive primes

    p_1,p_2,...,p_t

    p_t - p_(t-1) = K
    p_(t-1) - p_(t-2) = K
    .
    .
    .
    p_2 - p_1 = K

    From here, I've been told to consider the sum of these differences and I should arrive to a conclusion that should contradict the prime number theorem.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Go along with the hint. What do you get when you add both sides of your equations?

    Also, what is  t ?
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  3. #3
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    I think that t is supposed to N. After reading what I've written, I'm assuming the conclusion, which I don't want to do right?

    But, if I sum up those pairs I get

    p_N - p_1 = (N-1)K

    But I am not sure how that is supposed to get me to a contradiction of the prime number theorem?
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  4. #4
    Super Member Bacterius's Avatar
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    There is something wrong : you are actually using the PNT in a proof that should invalidate it ? I don't understand, the PNT has been proven and any attempt to contradict it must end in failure, right ?

    Could you please reformulate the question ?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Bacterius View Post
    There is something wrong : you are actually using the PNT in a proof that should invalidate it ? I don't understand, the PNT has been proven and any attempt to contradict it must end in failure, right ?

    Could you please reformulate the question ?
    I was having the same problem understanding too.
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  6. #6
    Member Haven's Avatar
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    From what I see of the problem, it seems that the OP is trying to assume that there are less than N or N pairs of sucessive primes and trying to show that this contradicts the prime number theorem, and therefore there must be at least N consecutive primes.

    The easiest way to get N pairs of consecutive primes is to have all the prime be consecutive and have the property that \forall i \in \{1\dots t\} p_{i+1} - p_i = k.

    However, I think this is assuming a little too much. If we assume that t<=N and \forall i \in \{1\dots t\} p_{i+1} - p_i = n. for some n, then we get that p_{t}-p_1 = (t-1)n

    If we assume that primes are greater than 2, then it should follow that n is even, which will be our k. Now, using the fact that t<=N, we should be able to derive a contradiction thus proving that t>N.

    However, we must first prove that such a list of consecutive primes exists.
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  7. #7
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    Is that you, Richard?

    Quote Originally Posted by Haven View Post
    However, we must first prove that such a list of consecutive primes exists.
    This claim. Check this link Primes in arithmetic progression - Wikipedia, the free encyclopedia

    There you will read that such lists of consecutive primes are conjectured to exist but there is no proof yet. So this approach is not going to work.

    The question actually means that we need to find more than N pairs of "successive primes" i.e.

    p(a), p(a+1) ---> p(a+1) - p(a) = K
    p(b), p(b+1) ---> p(b+1) - p(b) = K
    p(c), p(c+1) ---> p(c+1) - p(c) = K

    etc where a, b, c are arbitrary, distinct indices.

    Btw, I haven't solved the problem yet and it's due in the morning
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  8. #8
    Member Haven's Avatar
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    I actually now think the idea of the setup is to set t to be so large that we can apply the Prime Number Theorem. We can make t arbitrarily large, so that it can apply to any N.

    so \pi(p_t - p_1) \approx t\log{t}

    so assuming we have N or less pairs of primes with difference k, we have to form a bound on the differences, in order to contradict this result.
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Haven View Post
    I actually now think the idea of the setup is to set t to be so large that we can apply the Prime Number Theorem. We can make t arbitrarily large, so that it can apply to any N.

    so \pi(p_t - p_1) \approx t\log{t}

    so assuming we have N or less pairs of primes with difference k, we have to form a bound on the differences, in order to contradict this result.
    I'm not quite seeing the contradiction. What if the other prime's differences compensate?
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  10. #10
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    You can start by setting t to be arbitrary and consider the equations:

    p_{i+1} - p_{i} = k_{i} for [LaTeX ERROR: Convert failed] between 1 and t

    The values [LaTeX ERROR: Convert failed] are arbitrary integers and [LaTeX ERROR: Convert failed] are just the primes. For [LaTeX ERROR: Convert failed] 1, 2, 3,..., [LaTeX ERROR: Convert failed] 2, 3, 5,...

    Now what you can do is sum both sides of the equations. On the left you will get a telescoping sum and on the right you will just get the sum of the [LaTeX ERROR: Convert failed] 's. Then you need to think of what those values can be and bound them.

    I hope this helps.
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