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Math Help - TA’s Challenge Problem #2

  1. #1
    Senior Member TheAbstractionist's Avatar
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    TA’s Challenge Problem #2

    This is the second Challenge Problem I’ve made up.

    Prove that n^{4k}+4m^4 is not prime for all positive integers n,\,k,\,m with m>1.
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    Quote Originally Posted by TheAbstractionist View Post
    This is the second Challenge Problem I’ve made up.

    Prove that n^{4k}+4m^4 is not prime for all positive integers n,\,k,\,m with m>1.
    n^{4k}+4m^4=((n^k - m)^2 + m^2)((n^k + m)^2+m^2) and both factors are > 1 because m > 1.
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  3. #3
    Lord of certain Rings
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    NCA's identity has a name, its called Sophie Germain's Identity.
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    Senior Member TheAbstractionist's Avatar
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    Wow, that was really quick!
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    Grand Panjandrum
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    Quote Originally Posted by NonCommAlg View Post
    n^{4k}+4m^4=((n^k - m)^2 + m^2)((n^k + m)^2+m^2) and both factors are > 1 because m > 1.
    For real m and n and integer k:

    n^{4k}+4m^4=|n|^{4k}+4|m|^4

    So the restriction to positive integers for n and m is unnecessary.

    Also if m=0 we have n^{4k}+4m^4=n^{4k} which can never be prime for any integer n (including 0).

    So the problem can be rewordrd to for all positive integers k and integers n,m show that

    n^{4k}+4m^4

    is never prime.

    (we may have to be explicit about what we want 0^0 to mean here or exclude the case where one or both of n and m are zero and k is zero)

    CB
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    Quote Originally Posted by CaptainBlack View Post

    So the problem can be rewordrd to for all positive integers k and integers n,m show that

    n^{4k}+4m^4

    is never prime.
    this is not true. for example if n = m = 1, we'll get the prime number 5. it's true that m, n need not be positive though.
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    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    For real m and n and integer k:

    n^{4k}+4m^4=|n|^{4k}+4|m|^4

    So the restriction to positive integers for n and m is unnecessary.

    Also if m=0 we have n^{4k}+4m^4=n^{4k} which can never be prime for any integer n (including 0).

    So the problem can be rewordrd to for all positive integers k and integers n,m show that

    n^{4k}+4m^4

    is never prime.

    (we may have to be explicit about what we want 0^0 to mean here or exclude the case where one or both of n and m are zero and k is zero)

    CB
    That’s a very good point. However, the condition m>1 (or |m|>1 if you like) is absolutely vital, otherwise you could find a counterexample as NonCommAlg pointed out.

    The way I originally did it was to factorize n^{4k}+4m^4 as \left(n^{2k}+2mn^k+2m^2\right)\left(n^{2k}-2mn^k+2m^2\right) instead of writing the factors the way NonCommAlg did – and then it isn’t so clear that both factors are actually greater than 1. (One of them certainly is, but the other is iffy.) I then proceeded by considering the expressions n^{2k}\pm2mn^k+2m^2-1 as quadratics in n^k. Their determinant is 4m^2-8m^2+4=4(1-m^2) and the vital condition m>1 ensures that the determinant is negative and therefore that the expressions are positive.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by NonCommAlg View Post
    this is not true. for example if n = m = 1, we'll get the prime number 5. it's true that m, n need not be positive though.
    Aghh.. I had missed that |x|>1 excluded x=+/-1 and so had not considered those cases.

    CB
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