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Math Help - Euler's phi function

  1. #1
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    Euler's phi function

    I will denote the phi function to be &.
    For example, &(36) = 36(1-1/2)(1-1/3) = 12

    Find all n such that &(n) = 8.

    How is this done w/o just guessing and checking? Thanks...
    Last edited by jzellt; April 7th 2010 at 10:41 PM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by jzellt View Post
    I will denote the phi function to be &.
    For example, &(36) = 36(1-1/2)(1-1/3) = 12

    Find all n such that &(n) = 8. Find them and prove there is no more.

    How is this done w/o just guessing and checking? Thanks...
    \phi(n) = 8

    as you know

    \phi(p^n) = (p^n - p^{n-1})

    so n is a multiple of 2 just since if it is multiple from another prime it will be a divisor of phi(n) is it ok for now

    so n contains just 2

    suppose n = 2^t

    \phi(2^t) = 2^t - 2^{t-1} = 8

    2^{t-1}(2-1) = 8 \rightarrow t-1=3

    so

    n= 2^4
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  3. #3
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    I thought I understood, but I guess I'm still a bit confused...

    What if I had to find all n such that &(n) = 2?

    The above seems to calculate how many n, but not what each n is.

    Can you elaborate...

    Thanks.
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  4. #4
    MHF Contributor Amer's Avatar
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    "so n is a multiple of 2 just since if it is multiple from another prime it will be a divisor of phi(n) is it ok for now"

    this is wrong statement I forgot number theory sorry

    phi(5) = 4 and 2 is not a divisor of 5



    phi(15) = 2(4) = 8

    there is another solutions

    I can't figure exactly how I can solve it sorry
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Amer View Post
    \phi(n) = 8

    as you know

    \phi(p^n) = (p^n - p^{n-1})

    so n is a multiple of 2 just since if it is multiple from another prime it will be a divisor of phi(n) is it ok for now

    so n contains just 2

    suppose n = 2^t

    \phi(2^t) = 2^t - 2^{t-1} = 8

    2^{t-1}(2-1) = 8 \rightarrow t-1=3

    so

    n= 2^4
    The above works for when  n=p^a

    What if  n=p^a q^b ?

     \phi(n) = p^{a-1}(p-1)q^{b-1}(q-1) . Now if  p,q\neq2 we have then that  p^{a-1}=q^{b-1}=1 \implies a=b=1 and WLOG  p-1=4,\; q-1=2 \implies p=5,\; q=3 .
    Thus  n=15 .

    WLOG if  p=2 , then  \phi(n) = 2^{a-1}q^{b-1}(q-1) . This means again that  q^{b-1} = 1\implies b=1 . So we could have  q-1=2,\; 2^{a-1} = 4 or  q-1=4,\; 2^{a-1}=2 . Or in other words  q=3,\;a=3 or  q=5,\; a=2 .
    Thus  n=24 \text{ or } n=20

    What if  n=2p^aq^b ? I'll let you do this one (hint: what's  \phi(n) ?). The answer comes out to be  n=30 .

    Now what about other  n ? Well, it turns out that with three or more prime divisors, there are just too many factors in  \phi(n) i.e.  p_1-1,p_2-1,p_3-1,\cdots will all be even and distinct, thus making  \phi(n)>8 .


    So here's all  n with  \phi(n)=8 :  n=\{15,16,20,24,30\} .
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