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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
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    find all primes of the form  n^4 - m^4
    What???? Maybe, I am misinterpreting you, or I don't understand what's going on here, but if p=n^4-m^4 then p=\left(n^2+m^2\right)\left(n^2-m^2\right)
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Maybe the question is: Show there don't exist primes of the form n^4-m^4?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    Maybe the question is: Show there don't exist primes of the form n^4-m^4?
    That would be a pretty easy question.
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    MHF Contributor Drexel28's Avatar
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    There should be a subtle remark made here though. A prime is a natural number, so if p=n^4-m^4 we must have that n^4>m^4\implies n>m so that we know that n^2-m^2 is a positive divisor.
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  6. #6
    Senior Member Dinkydoe's Avatar
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    Aside from that: (n^2-m^2) \neq 1 for any n,m: n> m what would be required if p is prime.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Dinkydoe View Post
    Aside from that: (n^2-m^2) \neq 1 for any n,m: n> m what would be required if p is prime.
    You cannot have that n^2-m^2=1 since n^2-m^2=\left(n-m\right)\cdot\left(n+m\right) and since n+m>1 it follows that n-m<1.
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  8. #8
    Moo
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    I don't understand why you're debating oO
    So the answer's just : there's none !
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    I don't understand why you're debating oO
    So the answer's just : there's none !
    I agree. It was just two things had to resolved, clearly the factorization disproves that it can be prime unless n^2-m^2 is one or less than zero. Both of which were disproven.
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  10. #10
    Flow Master
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    I would prefer no more replies in this thread until the OP responds. Thankyou.
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