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Math Help - disproving prime numbers

  1. #1
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    disproving prime numbers

    how do you prove that number is not Prime?

    the example in my book says prove that 2 to the power of 66 is not prime and then says the keys lies in this simple algebra x squared-1=(x-1)(x+1) it says to use that to write a proof that the number is not prime so what am supposed to do with that equation?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by andrewuk View Post
    how do you prove that number is not Prime?

    the example in my book says prove that 2 to the power of 66 is not prime and then says the keys lies in this simple algebra x squared-1=(x-1)(x+1) it says to use that to write a proof that the number is not prime so what am supposed to do with that equation?
    huh
    2 divides 2^{66}, therefore it is not prime

    Another way :
    2^{66}-1=(2^{33}-1)(2^{33}+1)

    \implies 2^{66}=(2^{33}-1)(2^{33}+1)+1

    But since 2^{33} is an even number, 2^{33}-1 \text{ and } 2^{33}+1 are odd numbers. Thus their product is an odd number.
    If we substract 1 to it, we get an even number. Hence 2^{66} is an even number (and different from 2), so it is not prime.


    But that's finding the most complicated way...
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  3. #3
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    Ok thanks for that, don't know why the book couldn't of included an example first to show how the equation is used.

    How would i split it if the power is an odd number? e.g. 2 to the power of 67
    Last edited by andrewuk; September 27th 2008 at 04:23 AM. Reason: math code error
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    huh
    2 divides 2^{66}, therefore it is not prime

    Another way :
    2^{66}-1=(2^{33}-1)(2^{33}+1)

    \implies 2^{66}=(2^{33}-1)(2^{33}+1)+1

    But since 2^{33} is an even number, 2^{33}-1 \text{ and } 2^{33}+1 are odd numbers. Thus their product is an odd number.
    If we substract 1 to it, we get an even number. Hence 2^{66} is an even number (and different from 2), so it is not prime.


    But that's finding the most complicated way...
    This is ridiculus, the very definition of 2^{66} makes it composite and hence not prime as it has at least three distinct divisors. So any proof that 2^{66} is not prime that uses any of the laws of powers is using this, and so is just a complicated version of the simple proof.

    RonL
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  5. #5
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    Ye the example does seem stupid but i still need to know how to use that equation when the power is an odd number, can you have a decimal power?
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  6. #6
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    Its considered rude to post the same question on two forums.

    Prime numbers - The Student Room

    Bobak
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  7. #7
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    Given any number A with prime factorization x_0^{n_0} \cdot x_1^{n_1} \cdot \dots \cdot x_k^{n_k} is composite with (n_0 + 1) \cdot (n_1 + 1) \cdot \dots \cdot (n_k + 1) factors.

    So, this works for any prime factorization, whether it is to an odd power or not.

    I do not quite understand what you mean by decimal power. Do you mean can you have 2^{0.3}? If so, that number does exist, however you will not find factors for it.
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  8. #8
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    Hello,

    Here is my opinion: someone made an error when copying.
    The real question in the book should be "Prove that 2^{66}-1 is a prime."

    Bye.
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  9. #9
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    Quote Originally Posted by wisterville View Post
    Hello,

    Here is my opinion: someone made an error when copying.
    The real question in the book should be "Prove that 2^{66}-1 is a prime."

    Bye.
    I'm sorry now I'm lost.

    2^{66}-1 is not prime as:

    2^{66}-1=(2^{33}+1)(2^{33}-1)

    RonL
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  10. #10
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    Hello,

    Sorry, I wanted to say:
    The real question in the book should be "Prove that is NOT a prime."

    Bye.
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