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Math Help - Prime factorization phi question

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    Prime factorization phi question

    What is the 5th solution of ?

    This is of the form where \phi x = x (1 - 1/p1 ).... (1 - 1/pr)

    where p1....pr refer to the prime factorization of x
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sirellwood View Post
    What is the 5th solution of ?

    This is of the form where \phi x = x (1 - 1/p1 ).... (1 - 1/pr)

    where p1....pr refer to the prime factorization of x
    The 'equation' can be written as...

    (1)

    From (1) is evident that no prime factors >3 of x exist and 2 is not prime factor so that the only solution is x=3...

    Kind regards

    \chi \sigma
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    Thanks! yeah this is what i found last nite before i went to bed.... i found that 3 was an answer so mayb assumed it was numbers of the form 3^a prime factorization were answers or something :-s
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    iv just been informed that there are more answers than just 3.... and that there is a numerical solution to this question?
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sirellwood View Post
    iv just been informed that there are more answers than just 3.... and that there is a numerical solution to this question?
    ... effectively the equation...

    (1)

    ... indicates that 3 is the only prime factor of x and that means that any is solution of (1)...

    Kind regards

    \chi \sigma
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    ok maybe i am missing something fundamental here because thats what i assumed you had meant originally.... but if any 3^k is a solution, then take 3^2 = 9 for example....

    we get p1=3 and p2=3

    so (1) = (3-1)(3-1)/9 = 4/9 which does not equal 2/3 as required?
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sirellwood View Post
    ok maybe i am missing something fundamental here because thats what i assumed you had meant originally.... but if any 3^k is a solution, then take 3^2 = 9 for example....

    we get p1=3 and p2=3

    so (1) = (3-1)(3-1)/9 = 4/9 which does not equal 2/3 as required?
    The 'Fundamental Theorem of Arithmetic'm extablishes that any n>1 can be written as...

    (1)

    ... so that...

    (2)

    ... depends only from the 'prime factors' and not from the 'exponents' . In Your example is...



    ... because 9 and 3 have both the only prime factor 3...

    Kind regards

    \chi \sigma
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  8. #8
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    \varphi(p^k) is not k\varphi(p) = k(p-1), but rather (p-1)p^{k-1}
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