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Re: [PrimeNumbers] Prime formula...

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  • Andrew Swallow
    ... phi(n^2)=n*phi(n), for any n. so boils down to showing that phi(n)=n-1 iff n is prime. which is easily known. Andy
    Message 1 of 4 , May 3 12:01 PM
      > phi( (p^2) ) = [phi(p)]^2 + phi(p) iff "p" is prime,
      > and phi(...) is the totient function.
      >
      > Or is this equation already known? Bill.


      phi(n^2)=n*phi(n), for any n. so boils down to showing
      that phi(n)=n-1 iff n is prime. which is easily known.

      Andy





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    • Shiv Nandan Singh
      phi(p^2) = p*( p-1)......(1) phi(p)=p-1..........(2) Using 1 and 2 in phi( (p^2) ) = [phi(p)]^2 + phi(p) we get p*(p-1) = (p-1)^2 + p-1 I think the result is
      Message 2 of 4 , May 3 5:46 PM
        phi(p^2) = p*( p-1)......(1)
        phi(p)=p-1..........(2)

        Using 1 and 2
        in
        phi( (p^2) ) = [phi(p)]^2 + phi(p)
        we get
        p*(p-1) = (p-1)^2 + p-1

        I think the result is obvious.........for the if
        case.

        Now for the only if case

        Let N = p(1)^a(1) * p(2)^a92)*........*p(n)^a(n)

        phi(N) =
        p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}

        N^2= p(1)^2a(1) * p(2)^2a(2)*........*p(n)^2a(n)

        phi(N^2) =
        p(1)^{2a(1)-1}*p(2)^{2a(2)-1}*......p(n)^{2a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}

        Now if the equation holds then

        phi( (p^2) ) = [phi(p)]^2 + phi(p)

        phi(N^2) = phi(N)^2 + phi(N)
        ==>
        [p(1)^{2a(1)-1}*p(2)^{2a(2)-1}*......p(n)^{2a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}]
        =
        [p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}]^2
        +
        [p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}]

        Cancelling off the common terms from both the sides
        (implies none of p(i) = 1 or 0 )

        The equation yields

        p(1)^a(1) * p(2)^a92)*........*p(n)^a(n)=
        p(1)^{a(1)-1}*p(2)^{a(2)-1}*......p(n)^{a(n)-1}*{p(1)-1}*{p(2)-1}.....{p(n)-1}
        + 1

        N = phi(N) + 1

        Or phi(N) = N-1
        which is true iff N = prime


        I think the proof is over
        and the statement is very correct.

        Shiv Nandan Singh





        Shiv Nandan Singh

        --- William Bouris <melangebillyb@...> wrote:
        > Hello Group...
        >
        > Can someone prove or disprove?
        >
        > phi( (p^2) ) = [phi(p)]^2 + phi(p) iff "p" is prime,
        > and phi(...) is the totient function.
        >
        > Or is this equation already known? Bill.
        >
        >
        > -




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