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Math Help - number theory, solve diophantine equation

  1. #1
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    Exclamation number theory, solve diophantine equation

    Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.

    Thanks very much guys
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  2. #2
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    Quote Originally Posted by suedenation View Post
    Prove that if (2 to the power n) - 15 = x square, then n = 4 or n = 6.

    Thanks very much guys
    We need,
    2^n-15=x^2
    If,
    n=2k
    Then,
    (2^k)^2-15=x^2
    Thus,
    (2^k)^2-x^2=15
    Thus,
    (2^k-x)(2^k+x)=15
    The factorization of 15 is,
    1\cdot 15,3\cdot 5
    Examine each of the cases to get the result.
    Note,
    2^k-x<2^k+x
    Thus, we have only two possibilities,
    \left\{ \begin{array}{c}2^k-x=1\\2^k+x=15\end{array} \right\}
    \left\{ \begin{array}{c}2^k-x=3\\2^k+x=5\end{array} \right\}
    The equations respectively,
    2\cdot 2^k=2^{k+1}=16\to k+1=4\to k=3
    2\cdot 2^k=2^{k+1}=8\to k+1=3\to k=2
    In each case we have,
    n=2(3)=6
    n=2(2)=4
    ----
    If,
    n=1 then,
    2-15=x^2--->Impossible.
    If n>1 and
    n=2k+1
    Then,
    2(2^k)^2-15=x^2
    Thus,
    x^2-2(2^k)^2=-15
    This is a Pellian look-a-like equation.
    (Now I am hoping that it has no solution to it).
    I have to think about this diophantine equation.
    Basically I need to show,
    a^2-2b^2=-15
    Has no solutions.
    Last edited by ThePerfectHacker; December 3rd 2006 at 07:27 PM.
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  3. #3
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    Aha! Solved it.

    The diophantine equation,
    a^2-2b^2=-15
    Has no solution!

    The left hand side needs to be divisible by 15
    Thus,
    a^2-2b^2\equiv 0(\mbox{ mod }3)
    a^2-2b^2\equiv 0(\mbox{ mod }5)
    This means the Legendre symbols have the value,
    (2b^2/3)=(2/3)=1
    (2b^2/5)=(2/5)=1
    But that is not true!
    Because by Euler's criterion,
    2^{\frac{3-1}{2}}\equiv -1 (\mbox{ mod } 3)
    2^{\frac{5-1}{2}}\equiv -1 (\mbox{ mod }5)
    Last edited by ThePerfectHacker; December 3rd 2006 at 07:27 PM.
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  4. #4
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    Exclamation

    Thanks sooooo much, you are a genius....hehe
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