Results 1 to 4 of 4

Thread: number theory, solve diophantine equation

  1. #1
    Junior Member
    Joined
    Feb 2006
    From
    Canada
    Posts
    45

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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,
    $\displaystyle 2^n-15=x^2$
    If,
    $\displaystyle n=2k$
    Then,
    $\displaystyle (2^k)^2-15=x^2$
    Thus,
    $\displaystyle (2^k)^2-x^2=15$
    Thus,
    $\displaystyle (2^k-x)(2^k+x)=15$
    The factorization of 15 is,
    $\displaystyle 1\cdot 15,3\cdot 5$
    Examine each of the cases to get the result.
    Note,
    $\displaystyle 2^k-x<2^k+x$
    Thus, we have only two possibilities,
    $\displaystyle \left\{ \begin{array}{c}2^k-x=1\\2^k+x=15\end{array} \right\}$
    $\displaystyle \left\{ \begin{array}{c}2^k-x=3\\2^k+x=5\end{array} \right\}$
    The equations respectively,
    $\displaystyle 2\cdot 2^k=2^{k+1}=16\to k+1=4\to k=3$
    $\displaystyle 2\cdot 2^k=2^{k+1}=8\to k+1=3\to k=2$
    In each case we have,
    $\displaystyle n=2(3)=6$
    $\displaystyle n=2(2)=4$
    ----
    If,
    $\displaystyle n=1$ then,
    $\displaystyle 2-15=x^2$--->Impossible.
    If $\displaystyle n>1$ and
    $\displaystyle n=2k+1$
    Then,
    $\displaystyle 2(2^k)^2-15=x^2$
    Thus,
    $\displaystyle x^2-2(2^k)^2=-15$
    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,
    $\displaystyle a^2-2b^2=-15$
    Has no solutions.
    Last edited by ThePerfectHacker; Dec 3rd 2006 at 07:27 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Aha! Solved it.

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

    The left hand side needs to be divisible by 15
    Thus,
    $\displaystyle a^2-2b^2\equiv 0(\mbox{ mod }3)$
    $\displaystyle a^2-2b^2\equiv 0(\mbox{ mod }5)$
    This means the Legendre symbols have the value,
    $\displaystyle (2b^2/3)=(2/3)=1$
    $\displaystyle (2b^2/5)=(2/5)=1$
    But that is not true!
    Because by Euler's criterion,
    $\displaystyle 2^{\frac{3-1}{2}}\equiv -1 (\mbox{ mod } 3)$
    $\displaystyle 2^{\frac{5-1}{2}}\equiv -1 (\mbox{ mod }5)$
    Last edited by ThePerfectHacker; Dec 3rd 2006 at 07:27 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2006
    From
    Canada
    Posts
    45

    Exclamation

    Thanks sooooo much, you are a genius....hehe
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Textbooks on Galois Theory and Algebraic Number Theory
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jul 8th 2011, 06:09 PM
  2. Replies: 1
    Last Post: Mar 17th 2010, 01:43 PM
  3. How does one solve a quadratic Diophantine equation?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 17th 2009, 04:46 AM
  4. Replies: 2
    Last Post: Dec 18th 2008, 05:28 PM
  5. Number Theory - Primes and Eulers equation
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: Aug 9th 2007, 11:39 AM

Search Tags


/mathhelpforum @mathhelpforum