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Math Help - The function o(n)

  1. #1
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    The function o(n)

    Hello all, I have questions about the function o(n) as well. o(n) is the sum of the divisors of n including n itself.

    1a. First I need the find the n for which o(n) = 15, and then prove that n is unique. How do you accomplish this ?

    1b. Similarly, how can I prove that no solutions exist to the equation o(n) = 17?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    I don't have time to solve this symbolically but I will tell you this:

     \sigma(n)>n so we only have finitely many numbers to check.

    Scanning through we get  \sigma(8)=15 and no numbers satisfy  \sigma(n)=17 .
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    I'm sorry, but if someone can provide a little clearer and more thorough explanation for both of my questions, I'd really appreciate it!
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    I'm sorry, but if someone can provide a little clearer and more thorough explanation for both of my questions, I'd really appreciate it!
    What's unclear to you?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I don't have time to solve this symbolically but I will tell you this:

     \sigma(n)>n so we only have finitely many numbers to check.

    Scanning through we get  \sigma(8)=15 and no numbers satisfy  \sigma(n)=17 .
    What about this? Let n\in\mathbb{N} be arbitrary. Then, n=p_1^{\alpha_1}\cdots p_m^{\alpha_m} and so \sigma(n)=\prod_{j=1}^{m}\frac{p_m^{\alpha_m+1}-1}{p_m-1}. So, if \sigma(n)=17 it follows that 17=\frac{p_m^{\alpha_m+1}-1}{p-1} from it's factorization properties. But, this means that 17=1+\cdots+p_m^{\alpha_m} and thus 16=p_m^1+\cdots+p_m^{\alpha_m} but this means that 16=p_m(1+\cdots+p_m^{\alpha_m}) and thus p_m=2\implies 8=1+\cdots+p_m^{\alpha_m-1} which is ridicuoulous why?
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    What about this? Let n\in\mathbb{N} be arbitrary. Then, n=p_1^{\alpha_1}\cdots p_m^{\alpha_m} and so \sigma(n)=\prod_{j=1}^{m}\frac{p_m^{\alpha_m+1}-1}{p_m-1}. So, if \sigma(n)=17 it follows that 17=\frac{p_m^{\alpha_m+1}-1}{p-1} from it's factorization properties. But, this means that 17=1+\cdots+p_m^{\alpha_m} and thus 16=p_m^1+\cdots+p_m^{\alpha_m} but this means that 16=p_m(1+\cdots+p_m^{\alpha_m}) and thus p_m=2\implies 8=1+\cdots+p_m^{\alpha_m-1} which is ridicuoulous why?
    I believe this argument shows there is no  n such that  \sigma(n)=2^m+1 where  m>1 and  2^m+1 is prime.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I believe this argument shows there is no  n such that  \sigma(n)=2^m+1 where  m>1 and  2^m+1 is prime.
    Agreed!
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    For the other problem try this:

     15=3\cdot5 and the other solution shows there is no  n such that  \sigma(n)=5 .

    This means if  \sigma(n)=15 then  n must be a prime power, otherwise if  n=ab where  (a,b)=1 then  \sigma(n)=\sigma(a)\sigma(b)=3\cdot5  which means  \sigma(b)=5 , which is impossible.

    So  n=p^m a prime power means  15=p^m+\cdots+1 . Thus  14=p^m+\cdots+p=p\cdot(p^{m-1}+\cdots+1) which forces  p=2 or  7 . Next see that  p=2 is the only option. Now you can find the exponent fairly easily.

    This argument is a bit sloppy but I think I give enough info to piece it together.
    Last edited by chiph588@; May 28th 2010 at 03:58 PM.
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    What about this? Let n\in\mathbb{N} be arbitrary. Then, n=p_1^{\alpha_1}\cdots p_m^{\alpha_m} and so \sigma(n)=\prod_{j=1}^{m}\frac{p_m^{\alpha_m+1}-1}{p_m-1}. So, if \sigma(n)=17 it follows that 17=\frac{p_m^{\alpha_m+1}-1}{p-1} from it's factorization properties. But, this means that 17=1+\cdots+p_m^{\alpha_m} and thus 16=p_m^1+\cdots+p_m^{\alpha_m} but this means that 16=p_m(1+\cdots+p_m^{\alpha_m}) and thus p_m=2\implies 8=1+\cdots+p_m^{\alpha_m-1} which is ridicuoulous why?
    Why is that ridiculous? I'm failing to recognize this...
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by 1337h4x View Post
    Why is that ridiculous? I'm failing to recognize this...
    Well one side is even and the other side is odd.
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