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  1. #1
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    urgent

    can you please help?

    Find all twin primes p, p+2 whose mid term p+1 is i) Triangular ii) Perfect square iii) perfect


    thx
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  2. #2
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    Quote Originally Posted by marlen19861@hotmail.com View Post
    can you please help?

    Find all twin primes p, p+2 whose mid term p+1 is i) Triangular ii) Perfect square iii) perfect


    thx
    i) If $\displaystyle p+1$ is a triangular number there exists a natural number $\displaystyle k$ such that:

    $\displaystyle p+1=\frac{k(k+1)}{2}$

    so:

    $\displaystyle k^2+k-(2p+2)=0$

    and hence:

    $\displaystyle k=\frac{-1 \pm \sqrt{p+2}}{2}$

    But the left hand side is an integer and the right hand side is irrational if $\displaystyle p+2$ is not a perfect square, but $\displaystyle p+2$ is prime. Hence there exist no twin priimes whose mid term is a triangular number.

    CB
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  3. #3
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    is the last question correct?
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  4. #4
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    Quote Originally Posted by nikolany View Post
    is the last question correct?
    Now that I think about it it can't be since 5,6,7 would be a counter example.

    The last equation should be:

    $\displaystyle k=\frac{-1\pm\sqrt{8p+9}}{2}$

    Then we require that $\displaystyle 8p+9$ be a perfect square for both sides to be rational, which does not obviously work

    CB
    Last edited by CaptainBlack; Nov 10th 2008 at 12:20 AM.
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Now that I think about it it can't be since 5,6,7 would be a counter example.

    The last equation should be:

    $\displaystyle k=\frac{-1\pm\sqrt{8p+9}}{2}$

    Then we require that $\displaystyle 8p+9$ be a perfect square for both sides to be rational, which does not obviously work

    CB
    Having thought about this some more I can now make this work, but I won't go into the detail of the solution as Opalg's solution in the other thread is much neater.

    But the gist of the solution is that we can reduce the condition that $\displaystyle 8p+9$ is a perfect square to: there exists a positive integer $\displaystyle k$ such that

    $\displaystyle 2p=(k-1)(k+2)$

    which by the fundamental theorem of arithmetic forces us to conclude $\displaystyle p=??$.

    CB
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