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Math Help - The function d(n)

  1. #1
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    The function d(n)

    Hello everyone,

    I have some questions about the function d(n) which gives the number of positive divisors of n including n itself.

    1. How can it be proven that d(n) is odd IFF n is a perfect square? (Kind of like proving that d(n) is odd iff n = k^2 for some integer k.

    2. How is the following statement true: The product of all of the positive divisors of n (including n itself) is n^[d(n)/2] . Is there a proof for this ?

    Any help is appreciated!
    Last edited by 1337h4x; May 21st 2010 at 04:59 PM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    Hello everyone,

    I have some questions about the function d(n) which gives the number of positive divisors of n including n itself.

    1. How can it be proven that d(n) IFF n is a perfect square? (Kind of like proving that d(n) is odd iff n = k^2 for some integer k.

    2. How is the following statement true: The product of all of the positive divisors of n (including n itself) is n^[d(n)/2] . Is there a proof for this ?

    Any help is appreciated!
    1. That's not even a sentence! d(n) what if and only if (...)?

    2. Look at the power with which a prime dividing n appears in the canonical factorization of the resulting product.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    1. That's not even a sentence! d(n) what if and only if (...)?

    2. Look at the power with which a prime dividing n appears in the canonical factorization of the resulting product.
    I fixed it, It should say d(n) is odd IFF ...

    Anyways, what do you mean by your point of 2? Can you also explain my first question?
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  4. #4
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    Can anyone help with this? I'm struggling to understand this.
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  5. #5
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    Can anyone please explain the second question at least? I'm really stuck on this concept.
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    Super Member PaulRS's Avatar
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    For all parts, note that if d|n then \left(\tfrac{n}{d}\right)|n, the idea would be the following:

    For each divisor d<\sqrt{n} there's a corresponding one \tfrac{n}{d}>\sqrt{n}, the only divisor that is not counted here is d = \tfrac{n}{d}=\sqrt{n} in case of this being possible (that is, n is a square).

    Thus if n is not a square the divisors come in pairs, hence the number of divisors in that case is even, and in the case of n being a square it is odd (we add one).

    For the second part use the above idea and note that d\cdot \left(\tfrac{n}{d}\right)=n -and now try to use it to calculate the product of the divisors-
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    Quote Originally Posted by PaulRS View Post
    For all parts, note that if d|n then \left(\tfrac{n}{d}\right)|n, the idea would be the following:

    For each divisor d<\sqrt{n} there's a corresponding one \tfrac{n}{d}>\sqrt{n}, the only divisor that is not counted here is d = \tfrac{n}{d}=\sqrt{n} in case of this being possible (that is, n is a square).

    Thus if n is not a square the divisors come in pairs, hence the number of divisors in that case is even, and in the case of n being a square it is odd (we add one).

    For the second part use the above idea and note that d\cdot \left(\tfrac{n}{d}\right)=n -and now try to use it to calculate the product of the divisors-

    What happens if you calculate the product of the divisors? I can't get this to related to n^(d(n)/2)....
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    Wait, my problem involved d(n), your solution never mentioned d(n) it mentioned d...
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    Wait, my problem involved d(n), your solution never mentioned d(n) it mentioned d...
    His solution is adequate but look at it like this: I'm going to assume  n is not square.

    Your product =\prod_{d|n}d = \prod_{d|n, \; d<\sqrt{n}} d \left(\frac{n}{d}\right) = \prod_{d|n, \; d<\sqrt{n}} n = n^{d(n)/2}
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  10. #10
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    for 1) let n=\prod_{j=1}^k p_j^{r_j} be the prime factorization of n. then d(n)=\prod_{j=1}^k (1+r_j). clearly d(n) is odd if and only if r_j is even for all j, i.e. n is a perfect square.

    for 2) let P=\prod_{d \mid n} d. then we'll also have P = \prod_{d \mid n} \frac{n}{d} and thus P^2 = \prod_{d \mid n} d \cdot \frac{n}{d} = \prod_{d \mid n} n = n^{d(n)}.
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