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Math Help - primes...

  1. #1
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    primes...

    find all primes of the form  n^4 -m^4
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  2. #2
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    Quote Originally Posted by flower3 View Post
    find all primes of the form  n^4 -m^4

    n^4-m^4=(n-m)(n+m)(n^2+m^2) ...

    Tonio
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  3. #3
    Super Member Bacterius's Avatar
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    So, to give a little hint, how many primes can you think of that satisfy n^4-m^4=(n-m)(n+m)(n^2+m^2) ? And what condition can you put on n and m so that the expression is prime ? (remember the basic definition of a prime number) Conclude.
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  4. #4
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    There aren't any primes of the form a^4 - b^4

    To see this,

    let n = a^4 - b^4

    factor...

    ...
    ...

    and you get a factor of (a + b) in the product which is neither 1 nor n and certainly divides n.

    Done!

    Regards,

    Shahz.
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  5. #5
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    Quote Originally Posted by tonio View Post
    n^4-m^4=(n-m)(n+m)(n^2+m^2) ...

    Tonio
    Tonio pretty much nailed it.
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  6. #6
    Super Member Bacterius's Avatar
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    Quote Originally Posted by pollardrho06 View Post
    Tonio pretty much nailed it.
    What I wanted Flower3 to think about was that there "might" have been primes of that form, and I wanted him (or most likely her) to investigate how this could have been a prime. The observation should be that if two of the factors are equal to 1, then the number is prime. And the conclusion would be that no two factors can be simultaneously equal to one, hence there is no prime of that form ...

    I thought that was pedagogic.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    This thread takes me on a stroll down memory lane.
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  8. #8
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    Quote Originally Posted by Drexel28 View Post
    This thread takes me on a stroll down memory lane.


    Unbelievable and annoying: same poster, same question , 1.5 months apart. No doubt, some would be better off knitting handkerchiefs....really!

    Tonio
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  9. #9
    Super Member Bacterius's Avatar
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    It's actually the same guy!?
    Geez, now that's impressive...
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