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Math Help - perfect square

  1. #1
    Junior Member Singular's Avatar
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    perfect square

    Prove that there is only one value for n that make

    2^8 + 2^11 + 2^n is a perfect square.


    Here is my attempt :

    2^8 + 2^11 + 2^n = m^2
    2^n = m^2 - 2^8 - 2^11
    2^n = (m-48)(m+48)

    According to a theorem (I don't know what it is if somebody know please tell me) that there exist non-negative number s and t so that m - 48 = 2^s , m + 48 = 2^t , s+t=n


    So :

    2^s + 48 = 2^t -48
    2^s - 2^t = 96
    2^s{2^(t-s) - 1} = 2^5 x 3

    s = 5 and t = 7, so n= s+t = 12
    Is right or there is a better way ? thanks
    Last edited by Singular; April 21st 2007 at 10:11 PM. Reason: Typo
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  2. #2
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    Have you tried induction?
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  3. #3
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    Quote Originally Posted by Singular View Post
    Prove that there is only one value for n that make

    2^8 + 2^11 + 2^n is a perfect square.
    You started out well,

    2^n + 2^8 + 2^11 = m^2

    2^n = m^2 - 48^2

    2^n = (m+48)(m-48)

    So, by the Fundamental Theorem of Arithmetic, (mentioned in Elements, Book IX Proposition XIV).

    m+48 = 2^s
    m-48 = 2^t
    s+t=n

    And your solution is perfect.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Singular View Post
    m^2 - 48 = 2^s , m^2 + 48 = 2^t , s+t=n
    I was confused on where you got this, but I see now that this is just a typo. (I figured it was, but I wasn't sure.)
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  5. #5
    Junior Member Singular's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I was confused on where you got this, but I see now that this is just a typo. (I figured it was, but I wasn't sure.)
    Sorry it's typo
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