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Math Help - Primes

  1. #1
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    Primes

    How could you prove that if a^n-3^n is a positive prime number, then a=4 and n is a positive prime number? (all numbers are natural)

    Am I supposed to try and factor a^n-b^n to see for which a and b the factorization is possible and for which it's not? I am having trouble with that factorization anyhow, so does anyone know how to approach this problem?

    I mean it seems to work for small n, but I just can not figure out how to do it for all natural numbers n...

    Thankful for answer!
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Primes

    Consider what happens if n is not a prime number, so n = pq. You can show that a^{pq} - b^{pq} is divisible by  a^p - b^p. Specifically:

     a^{pq} - b^{pq} = (a^p-b^p)(a^{p(q-1)} + a^{p(q-2)}b^p + a^{p(q-3)}b^{2p} + ..+a^pb^{(q-2)p} + b^{(q-1)p}

    Therefore a^{pq} - 3^{pq} cannot be prime unless a^p-3^p = 1, and that can only happen if a = 4 and p=1.
    Last edited by ebaines; September 16th 2011 at 11:11 AM.
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    Re: Primes

    Thanks for such a quick answer! That helps alot, thanks.
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  4. #4
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    Re: Primes

    Hi again!

    A follow-up question on the problem above. Is the following true:

    a^(pq)-b^(pq)=(a^p)^q-(b^p)^q=(a^p-b^p)(...)

    If yes: how can you know that for sure?
    If no: how do you go from q^(pq)-b^(pq) to (a^p-b^p)(...)

    Thanks!
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Primes

    You can divide  a^p - b^p into  a^{pq}-b^{pq} using long division to get the result that I showed earlier. And yes, it is true that
     a^{pq} - b^{pq} = (a^p)^q-(b^p)^q = (a^p-b^p)(...) , but you still need to use long division to find out what goes in the (...) part.
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    Re: Primes

    Quote Originally Posted by ebaines View Post
    You can divide  a^p - b^p into  a^{pq}-b^{pq} using long division to get the result that I showed earlier. And yes, it is true that
     a^{pq} - b^{pq} = (a^p)^q-(b^p)^q = (a^p-b^p)(...) , but you still need to use long division to find out what goes in the (...) part.
    Sorry I was being unclear. I understand that it is true, but is it enough to say that a^{pq} - b^{pq} = (a^p)^q-(b^p)^q to clarify that (a^p-b^p) is in fact a factor? Since I'm unsure I think it's not ^^. Could you maybe explain a bit further in what way the rewriting (a^p)^q-(b^p)^q leads to the answer? Unless it's not done by that rewriting ^^

    /Pkaff
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  7. #7
    MHF Contributor ebaines's Avatar
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    Re: Primes

    Quote Originally Posted by pkaff View Post
    Sorry I was being unclear. I understand that it is true, but is it enough to say that a^{pq} - b^{pq} = (a^p)^q-(b^p)^q to clarify that (a^p-b^p) is in fact a factor?
    No - it's not automatically obvious (at least not to me).

    Quote Originally Posted by pkaff View Post
    Since I'm unsure I think it's not ^^. Could you maybe explain a bit further in what way the rewriting (a^p)^q-(b^p)^q leads to the answer? Unless it's not done by that rewriting ^^
    You can show that  x^q - y^q = (x-y)(x^{q-1} +x^{q-2}y + x^{q-2}y^2 + ...+ y^{q-1}). So replace x by  a^p and y by  b^p to get the result.
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  8. #8
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    Re: Primes

    Quote Originally Posted by ebaines View Post
    So replace x by  a^p and y by  b^p to get the result.
    Lol obviously xP... I already did the x^n-y^n thingy, and if you replace x with x^p it will look like (x^p)^n, aswell as x^p instead of "x". I'm rambeling, thanks for the help!
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