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Math Help - Sum of reciprocals of phi(p)

  1. #1
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    Sum of reciprocals of phi(p)

    Write \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{p-1} = \frac{a}{b}, with gcd(a,b)=1. Show that p^2 divides a if p \geq 5

    I have no clue how to express it as \frac{a}{b}.
    Last edited by comssa; November 8th 2009 at 10:32 AM.
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  2. #2
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    Quote Originally Posted by comssa View Post
    Write \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{p-1} = \frac{a}{b}, with gcd(a,b)=1. Show that p^2 divides a if p \geq 5

    I see that \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{p-1} = \frac{1}{\phi(2)} + \frac{1}{\phi(3)} + ... + \frac{1}{\phi(p)}
    but I have no clue how to express it as \frac{a}{b}.

    I can't understand what you think you can see here: 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{p-1}\neq \frac{1}{\phi(2)}+\frac{1}{\phi(3)}+\frac{1}{\phi(  4)}+...+\frac{1}{\phi(p)} , since, for example, \phi(4)=2\neq 3.

    Anyway, and since p>2 , for simplicity of the proof we'll prove that in fact 2\left(1+\frac{1}{2}+...+\frac{1}{p-1}\right)\equiv 0\!\!\!\pmod {p^2}

    Now, we can form pairs \frac{1}{i}\,,\,\frac{1}{p-i}\,,\,1\leq i\leq p-1\,\,and\,\,\frac{1}{i}+\frac{1}{p-i}=\frac{p}{i(p-i)} , so 2\left(1+\frac{1}{2}+...+\frac{1}{p-1}\right)=2\left(\frac{p}{1(p-1)}+\frac{p}{2(p-2)}+...+\frac{p}{\left(\frac{p-1}{2}\right)\left(\frac{p+1}{2}\right)}\right)=

    =p\sum\limits_{i=1}^{p-1}\frac{1}{i(p-i)}=p\sum\limits_{i=1}^{p-1}\frac{1}{-i^2}\,\;since\,\;p-i\equiv -i\!\!\!\pmod p , so we have to prove this last sum is divisible by p (the minus sign never minds: we can take it out of the sum).

    But \sum\limits_{i=1}^{p-1}\frac{1}{i^2}=\sum\limits_{i=1}^{p-1}i^2\!\!\!\pmod p as the only elements in \mathbb{Z}_p^{*} which are inverses to themselves are 1 and -1 and without these two we're left with an even number of

    elements, from 2 to p-2 (and here enters the assumption p\geq 5!), where we can pair each one with its inverse. Now, the well-known formula

    \sum\limits_{i=1}^{p-1}i^2=\frac{(p+1)(2p+1)}{6}p\, means the sum equals zero modulo p since, again, p is prime greater than 5. Q.E.D.

    Tonio
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