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Math Help - Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^n)− 1

  1. #1
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    Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^m)− 1

    Working through a question i got to this stage where i need to:

    Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^m)− 1

    Should i be looking to expand ((a^2)^n) + 1?
    Last edited by sirellwood; November 5th 2009 at 01:17 PM.
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  2. #2
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    Quote Originally Posted by sirellwood View Post
    Working through a question i got to this stage where i need to:

    Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^n)− 1

    Should i be looking to expand ((a^2)^n) + 1?

    You realize there is no m in your mathematical expressions, right?

    Tonio
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    argh sorry. my bad. edited :-)
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  4. #4
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    Quote Originally Posted by sirellwood View Post
    Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^m)− 1
    If a=2, n=2 and m=3 then a^{2n}+1 = 2^4+1= 17, a^{2m}-1=2^6-1=63 = 3^2\times 7 but 17 does not divide 63.
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  5. #5
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    Exponents not Associative

    Sirellwood:

    The operator "^" is NOT associative, that is, a^(2^n) does not equal (a^2)^n in general -- in Latex, a^{2^n}\neq (a^2)^n

    Theorem: For m>n, a^{2^n}+1|a^{2^m}-1
    Proof: a^{2^m}-1=(a^{2^{m-1}}+1)(a^{2^{m-1}}-1)=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1)...
    Continue factoring the successive differences of squares k times until m-k=n. QED

    Your theorem is quite true, if rendered to the page in proper notation.
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  6. #6
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    Quote Originally Posted by sirellwood View Post
    argh sorry. my bad. edited :-)

    I'm almost sure that you must require quite a bit more, namely: not only m > n but in fact n | m...check this.

    Tonio
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    Quote Originally Posted by Media_Man View Post
    Sirellwood:

    The operator "^" is NOT associative, that is, a^(2^n) does not equal (a^2)^n in general -- in Latex, a^{2^n}\neq (a^2)^n

    Theorem: For m>n, a^{2^n}+1|a^{2^m}-1
    Proof: a^{2^m}-1=(a^{2^{m-1}}+1)(a^{2^{m-1}}-1)=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1)...
    Continue factoring the successive differences of squares k times until m-k=n. QED

    Your theorem is quite true, if rendered to the page in proper notation.

    I supposed that if the OP had wanted a^{2^n} he could have writte a^(2^n) and not (a^2)^n, which means (a^2)^n=a^{2n}. OTOH, he could have simply written a^(2n) and thus I think you're right and he meant a^{2^n}.
    Even using simple ASCII one must be careful or even describe things by words when mathematical notation is cumbersome.

    Tonio
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  8. #8
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    On language

    one must be careful or even describe things by words when mathematical notation is cumbersome.
    Funny, if you read pre-twentieth century math books and proofs, they rarely rely on notation and speak almost solely in plain english words. "One more than an integer raised to a power of two must evenly divide one less than that same number raised any higher power of two..." Mathematical notation is a language unto itself that has extremely strict rules of grammar.
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