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Lehmer's Totient Problem Solved

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  • leavemsg1
    a copy of this proof can be found at www.oddperfectnumbers.com ... D. H. Lehmer s (1932) Totient Conjecture Stated: phi(n) divides (n -1) iff n is prime.
    Message 1 of 8 , May 23 7:02 PM
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      a copy of this proof can be found at www.oddperfectnumbers.com
      ...
      D. H. Lehmer's (1932) Totient Conjecture Stated:
      phi(n) divides (n -1) iff 'n' is prime.
      (proof, backward)
      if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
      divides (n -1), exactly.
      (proof, forward)
      let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
      let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
      other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
      = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
      / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
      be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
      1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
      / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
      [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
      cannot be more co-prime divisors of C than there are the total number
      of possible divisors! thus, by contradiction, 'n' must be prime.
      ...
      it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
      marker on the back of a receipt for car repairs. at first glance, it
      looks as though the problem needs to be solved using Fermat's little
      theorem; that is DEFINITELY NOT the case. enjoy!
      ...
      *QED
      05/21/2013
      ...
      Bill Bouris
      www.oddperfectnumbers.com
    • Jose Ramón Brox
      Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be that C=bC . Regards, Jose Brox 2013/5/24 leavemsg1 ... -- La
      Message 2 of 8 , May 24 3:43 AM
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        Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
        that C=bC'.

        Regards,
        Jose Brox


        2013/5/24 leavemsg1 <leavemsg1@...>

        > **
        >
        >
        > a copy of this proof can be found at www.oddperfectnumbers.com
        > ...
        > D. H. Lehmer's (1932) Totient Conjecture Stated:
        > phi(n) divides (n -1) iff 'n' is prime.
        > (proof, backward)
        > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
        > divides (n -1), exactly.
        > (proof, forward)
        > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
        > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
        > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
        > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
        > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
        > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
        > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
        > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
        > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
        > cannot be more co-prime divisors of C than there are the total number
        > of possible divisors! thus, by contradiction, 'n' must be prime.
        > ...
        > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
        > marker on the back of a receipt for car repairs. at first glance, it
        > looks as though the problem needs to be solved using Fermat's little
        > theorem; that is DEFINITELY NOT the case. enjoy!
        > ...
        > *QED
        > 05/21/2013
        > ...
        > Bill Bouris
        > www.oddperfectnumbers.com
        >
        >
        >



        --
        La verdad (blog de raciocinio pol�tico e informaci�n
        social)<http://josebrox.blogspot.com/>


        [Non-text portions of this message have been removed]
      • leavemsg1
        I ve presented to you the basics of how the proof would take place. If you d like to work through a few details to improve upon my idea, then do so. If you
        Message 3 of 8 , May 24 4:34 PM
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          I've presented to you the basics of how the proof would
          take place. If you'd like to work through a few details
          to improve upon my idea, then do so. If you aren't part
          of the solution, then you're part of the problem...
          rewards, Bill

          --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@...> wrote:
          >
          > Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
          > that C=bC'.
          >
          > Regards,
          > Jose Brox
          >
          >
          > 2013/5/24 leavemsg1 <leavemsg1@...>
          >
          > > **
          > >
          > >
          > > a copy of this proof can be found at www.oddperfectnumbers.com
          > > ...
          > > D. H. Lehmer's (1932) Totient Conjecture Stated:
          > > phi(n) divides (n -1) iff 'n' is prime.
          > > (proof, backward)
          > > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
          > > divides (n -1), exactly.
          > > (proof, forward)
          > > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
          > > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
          > > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
          > > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
          > > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
          > > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
          > > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
          > > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
          > > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
          > > cannot be more co-prime divisors of C than there are the total number
          > > of possible divisors! thus, by contradiction, 'n' must be prime.
          > > ...
          > > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
          > > marker on the back of a receipt for car repairs. at first glance, it
          > > looks as though the problem needs to be solved using Fermat's little
          > > theorem; that is DEFINITELY NOT the case. enjoy!
          > > ...
          > > *QED
          > > 05/21/2013
          > > ...
          > > Bill Bouris
          > > www.oddperfectnumbers.com
          > >
          > >
          > >
          >
          >
          >
          > --
          > La verdad (blog de raciocinio político e información
          > social)<http://josebrox.blogspot.com/>
          >
          >
          > [Non-text portions of this message have been removed]
          >
        • leavemsg1
          here s what the missing portion would be similar to... let k 1, n= b^j*C is composite, and k*phi(b^j*C)= b^j*C -1 where gcd(b, C)= 1. then we d find that
          Message 4 of 8 , May 25 8:30 AM
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            here's what the missing portion would be similar to... let
            k <> 1, n= b^j*C is composite, and k*phi(b^j*C)= b^j*C -1
            where gcd(b, C)= 1. then we'd find that phi(C)= [b^j*C -1]
            / [k*b^(j-1)*(b- 1) -1]... forcing us to realize that k= 1
            and phi((b-1)/b) = phi(C) which is still a contradiction!
            we don't have to doubt the eventual answer.(5 more minutes)
            www.oddperfectnumbers.com

            --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:
            >
            > I've presented to you the basics of how the proof would
            > take place. If you'd like to work through a few details
            > to improve upon my idea, then do so. If you aren't part
            > of the solution, then you're part of the problem...
            > rewards, Bill
            >
            > --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@> wrote:
            > >
            > > Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
            > > that C=bC'.
            > >
            > > Regards,
            > > Jose Brox
            > >
            > >
            > > 2013/5/24 leavemsg1 <leavemsg1@>
            > >
            > > > **
            > > >
            > > >
            > > > a copy of this proof can be found at www.oddperfectnumbers.com
            > > > ...
            > > > D. H. Lehmer's (1932) Totient Conjecture Stated:
            > > > phi(n) divides (n -1) iff 'n' is prime.
            > > > (proof, backward)
            > > > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
            > > > divides (n -1), exactly.
            > > > (proof, forward)
            > > > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
            > > > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
            > > > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
            > > > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
            > > > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
            > > > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
            > > > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
            > > > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
            > > > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
            > > > cannot be more co-prime divisors of C than there are the total number
            > > > of possible divisors! thus, by contradiction, 'n' must be prime.
            > > > ...
            > > > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
            > > > marker on the back of a receipt for car repairs. at first glance, it
            > > > looks as though the problem needs to be solved using Fermat's little
            > > > theorem; that is DEFINITELY NOT the case. enjoy!
            > > > ...
            > > > *QED
            > > > 05/21/2013
            > > > ...
            > > > Bill Bouris
            > > > www.oddperfectnumbers.com
            > > >
            > > >
            > > >
            > >
            > >
            > >
            > > --
            > > La verdad (blog de raciocinio político e información
            > > social)<http://josebrox.blogspot.com/>
            > >
            > >
            > > [Non-text portions of this message have been removed]
            > >
            >
          • leavemsg1
            ... D. H. Lehmer s (1932) Totient Conjecture Stated: phi(n) divides (n -1) iff n is prime. (proof, backward) if n is prime, then phi(n) = (n -1) by
            Message 5 of 8 , May 25 6:40 PM
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              ...
              D. H. Lehmer's (1932) Totient Conjecture Stated:
              phi(n) divides (n -1) iff 'n' is prime.
              (proof, backward)
              if 'n' is prime, then phi(n) = (n -1) by definition, and
              phi(n) | (n -1), exactly as desired.
              (proof, forward)
              let phi(n) divide (n -1), or k*phi(n)= (n -1), and assume
              'n' is composite, or n= b^j*C where gcd(b^j, C)= 1 and k <> 1.
              so, k*phi(b^j*C) implies k*b^j*C -k*b^(j-1)*C -1= b^j*C -1 and
              adding one to... and dividing by b^(j-1) on... both sides, and
              rearranging terms, we have... k= bC / [bC -1]. therefore, k= 1,
              and phi(b^j*C) = b^j*C -1, and b^j*C -b^(j-1)*C = b^j*C -1 im-
              plies b^(j-1)*C = 1, or b= C= 1, and 'n' isn't composite. thus,
              'n' must be prime. enjoy!
              ...
              *QED
              05/25/2013
              ...
              thanks, Jose B.; it's done! rewards, Bill

              --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:
              >
              > here's what the missing portion would be similar to... let
              > k <> 1, n= b^j*C is composite, and k*phi(b^j*C)= b^j*C -1
              > where gcd(b, C)= 1. then we'd find that phi(C)= [b^j*C -1]
              > / [k*b^(j-1)*(b- 1) -1]... forcing us to realize that k= 1
              > and phi((b-1)/b) = phi(C) which is still a contradiction!
              > we don't have to doubt the eventual answer.(5 more minutes)
              > www.oddperfectnumbers.com
              >
              > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
              > >
              > > I've presented to you the basics of how the proof would
              > > take place. If you'd like to work through a few details
              > > to improve upon my idea, then do so. If you aren't part
              > > of the solution, then you're part of the problem...
              > > rewards, Bill
              > >
              > > --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@> wrote:
              > > >
              > > > Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
              > > > that C=bC'.
              > > >
              > > > Regards,
              > > > Jose Brox
              > > >
              > > >
              > > > 2013/5/24 leavemsg1 <leavemsg1@>
              > > >
              > > > > **
              > > > >
              > > > >
              > > > > a copy of this proof can be found at www.oddperfectnumbers.com
              > > > > ...
              > > > > D. H. Lehmer's (1932) Totient Conjecture Stated:
              > > > > phi(n) divides (n -1) iff 'n' is prime.
              > > > > (proof, backward)
              > > > > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
              > > > > divides (n -1), exactly.
              > > > > (proof, forward)
              > > > > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
              > > > > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
              > > > > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
              > > > > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
              > > > > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
              > > > > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
              > > > > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
              > > > > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
              > > > > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
              > > > > cannot be more co-prime divisors of C than there are the total number
              > > > > of possible divisors! thus, by contradiction, 'n' must be prime.
              > > > > ...
              > > > > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
              > > > > marker on the back of a receipt for car repairs. at first glance, it
              > > > > looks as though the problem needs to be solved using Fermat's little
              > > > > theorem; that is DEFINITELY NOT the case. enjoy!
              > > > > ...
              > > > > *QED
              > > > > 05/21/2013
              > > > > ...
              > > > > Bill Bouris
              > > > > www.oddperfectnumbers.com
              > > > >
              > > > >
              > > > >
              > > >
              > > >
              > > >
              > > > --
              > > > La verdad (blog de raciocinio político e información
              > > > social)<http://josebrox.blogspot.com/>
              > > >
              > > >
              > > > [Non-text portions of this message have been removed]
              > > >
              > >
              >
            • leavemsg1
              the third time s a charm... I had to re-write the proof. now, it truly IS a classic. it s done, done, done! ... D. H. Lehmer s (1932) Totient Conjecture
              Message 6 of 8 , May 26 5:28 AM
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                the third time's a charm... I had to re-write the proof.
                now, it truly IS a classic. it's done, done, done!
                ...
                D. H. Lehmer's (1932) Totient Conjecture Stated: phi(n)
                divides (n -1) iff 'n' is prime.
                (proof, backward)
                if 'n' is prime, then phi(n) = (n -1) by definition, and
                phi(n) | (n -1), exactly as desired.
                (proof, forward)
                let phi(n) divide (n -1), such that k*phi(n)= (n -1), and
                assume 'n' is composite, n= b^j*C where gcd(b^j, C)= 1 and
                k <> 1. so, k*phi(b^j*C) = b^j*C -1, k*phi(b^j)*phi(C)= b^j
                *C -1, (k*b^j -k*b^(j-1))*phi(C)= b^j*C -1, or... phi(C)=
                [b^j*C -1] / [k*b^j - k*b^(j-1)] implies that k= b^(j-1)= 1.
                hence, phi(C) = (bC -1) / (b -1), b= C, and phi(C)= C +1.
                but, 'C' cannot have more co-prime divisors than it has a
                total number of divisors. thus, 'n' must be prime. enjoy!
                ...
                *QED
                05/26/2013
                ...

                --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:
                >
                > ...
                > D. H. Lehmer's (1932) Totient Conjecture Stated:
                > phi(n) divides (n -1) iff 'n' is prime.
                > (proof, backward)
                > if 'n' is prime, then phi(n) = (n -1) by definition, and
                > phi(n) | (n -1), exactly as desired.
                > (proof, forward)
                > let phi(n) divide (n -1), or k*phi(n)= (n -1), and assume
                > 'n' is composite, or n= b^j*C where gcd(b^j, C)= 1 and k <> 1.
                > so, k*phi(b^j*C) implies k*b^j*C -k*b^(j-1)*C -1= b^j*C -1 and
                > adding one to... and dividing by b^(j-1) on... both sides, and
                > rearranging terms, we have... k= bC / [bC -1]. therefore, k= 1,
                > and phi(b^j*C) = b^j*C -1, and b^j*C -b^(j-1)*C = b^j*C -1 im-
                > plies b^(j-1)*C = 1, or b= C= 1, and 'n' isn't composite. thus,
                > 'n' must be prime. enjoy!
                > ...
                > *QED
                > 05/25/2013
                > ...
                > thanks, Jose B.; it's done! rewards, Bill
                >
                > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                > >
                > > here's what the missing portion would be similar to... let
                > > k <> 1, n= b^j*C is composite, and k*phi(b^j*C)= b^j*C -1
                > > where gcd(b, C)= 1. then we'd find that phi(C)= [b^j*C -1]
                > > / [k*b^(j-1)*(b- 1) -1]... forcing us to realize that k= 1
                > > and phi((b-1)/b) = phi(C) which is still a contradiction!
                > > we don't have to doubt the eventual answer.(5 more minutes)
                > > www.oddperfectnumbers.com
                > >
                > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                > > >
                > > > I've presented to you the basics of how the proof would
                > > > take place. If you'd like to work through a few details
                > > > to improve upon my idea, then do so. If you aren't part
                > > > of the solution, then you're part of the problem...
                > > > rewards, Bill
                > > >
                > > > --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@> wrote:
                > > > >
                > > > > Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
                > > > > that C=bC'.
                > > > >
                > > > > Regards,
                > > > > Jose Brox
                > > > >
                > > > >
                > > > > 2013/5/24 leavemsg1 <leavemsg1@>
                > > > >
                > > > > > **
                > > > > >
                > > > > >
                > > > > > a copy of this proof can be found at www.oddperfectnumbers.com
                > > > > > ...
                > > > > > D. H. Lehmer's (1932) Totient Conjecture Stated:
                > > > > > phi(n) divides (n -1) iff 'n' is prime.
                > > > > > (proof, backward)
                > > > > > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
                > > > > > divides (n -1), exactly.
                > > > > > (proof, forward)
                > > > > > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
                > > > > > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
                > > > > > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
                > > > > > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
                > > > > > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
                > > > > > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
                > > > > > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
                > > > > > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
                > > > > > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
                > > > > > cannot be more co-prime divisors of C than there are the total number
                > > > > > of possible divisors! thus, by contradiction, 'n' must be prime.
                > > > > > ...
                > > > > > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
                > > > > > marker on the back of a receipt for car repairs. at first glance, it
                > > > > > looks as though the problem needs to be solved using Fermat's little
                > > > > > theorem; that is DEFINITELY NOT the case. enjoy!
                > > > > > ...
                > > > > > *QED
                > > > > > 05/21/2013
                > > > > > ...
                > > > > > Bill Bouris
                > > > > > www.oddperfectnumbers.com
                > > > > >
                > > > > >
                > > > > >
                > > > >
                > > > >
                > > > >
                > > > > --
                > > > > La verdad (blog de raciocinio político e información
                > > > > social)<http://josebrox.blogspot.com/>
                > > > >
                > > > >
                > > > > [Non-text portions of this message have been removed]
                > > > >
                > > >
                > >
                >
              • leavemsg1
                i forgot to add the word possible divisors.
                Message 7 of 8 , May 26 6:16 AM
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                  i forgot to add the word 'possible' divisors.

                  --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:
                  >
                  > the third time's a charm... I had to re-write the proof.
                  > now, it truly IS a classic. it's done, done, done!
                  > ...
                  > D. H. Lehmer's (1932) Totient Conjecture Stated: phi(n)
                  > divides (n -1) iff 'n' is prime.
                  > (proof, backward)
                  > if 'n' is prime, then phi(n) = (n -1) by definition, and
                  > phi(n) | (n -1), exactly as desired.
                  > (proof, forward)
                  > let phi(n) divide (n -1), such that k*phi(n)= (n -1), and
                  > assume 'n' is composite, n= b^j*C where gcd(b^j, C)= 1 and
                  > k <> 1. so, k*phi(b^j*C) = b^j*C -1, k*phi(b^j)*phi(C)= b^j
                  > *C -1, (k*b^j -k*b^(j-1))*phi(C)= b^j*C -1, or... phi(C)=
                  > [b^j*C -1] / [k*b^j - k*b^(j-1)] implies that k= b^(j-1)= 1.
                  > hence, phi(C) = (bC -1) / (b -1), b= C, and phi(C)= C +1.
                  > but, 'C' cannot have more co-prime divisors than it has a
                  > total number of /*possible*/ divisors. thus, 'n' must be
                  > prime.enjoy!
                  > ...
                  > *QED
                  > 05/26/2013
                  > ...
                  >
                  > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                  > >
                  > > ...
                  > > D. H. Lehmer's (1932) Totient Conjecture Stated:
                  > > phi(n) divides (n -1) iff 'n' is prime.
                  > > (proof, backward)
                  > > if 'n' is prime, then phi(n) = (n -1) by definition, and
                  > > phi(n) | (n -1), exactly as desired.
                  > > (proof, forward)
                  > > let phi(n) divide (n -1), or k*phi(n)= (n -1), and assume
                  > > 'n' is composite, or n= b^j*C where gcd(b^j, C)= 1 and k <> 1.
                  > > so, k*phi(b^j*C) implies k*b^j*C -k*b^(j-1)*C -1= b^j*C -1 and
                  > > adding one to... and dividing by b^(j-1) on... both sides, and
                  > > rearranging terms, we have... k= bC / [bC -1]. therefore, k= 1,
                  > > and phi(b^j*C) = b^j*C -1, and b^j*C -b^(j-1)*C = b^j*C -1 im-
                  > > plies b^(j-1)*C = 1, or b= C= 1, and 'n' isn't composite. thus,
                  > > 'n' must be prime. enjoy!
                  > > ...
                  > > *QED
                  > > 05/25/2013
                  > > ...
                  > > thanks, Jose B.; it's done! rewards, Bill
                  > >
                  > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                  > > >
                  > > > here's what the missing portion would be similar to... let
                  > > > k <> 1, n= b^j*C is composite, and k*phi(b^j*C)= b^j*C -1
                  > > > where gcd(b, C)= 1. then we'd find that phi(C)= [b^j*C -1]
                  > > > / [k*b^(j-1)*(b- 1) -1]... forcing us to realize that k= 1
                  > > > and phi((b-1)/b) = phi(C) which is still a contradiction!
                  > > > we don't have to doubt the eventual answer.(5 more minutes)
                  > > > www.oddperfectnumbers.com
                  > > >
                  > > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                  > > > >
                  > > > > I've presented to you the basics of how the proof would
                  > > > > take place. If you'd like to work through a few details
                  > > > > to improve upon my idea, then do so. If you aren't part
                  > > > > of the solution, then you're part of the problem...
                  > > > > rewards, Bill
                  > > > >
                  > > > > --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@> wrote:
                  > > > > >
                  > > > > > Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
                  > > > > > that C=bC'.
                  > > > > >
                  > > > > > Regards,
                  > > > > > Jose Brox
                  > > > > >
                  > > > > >
                  > > > > > 2013/5/24 leavemsg1 <leavemsg1@>
                  > > > > >
                  > > > > > > **
                  > > > > > >
                  > > > > > >
                  > > > > > > a copy of this proof can be found at www.oddperfectnumbers.com
                  > > > > > > ...
                  > > > > > > D. H. Lehmer's (1932) Totient Conjecture Stated:
                  > > > > > > phi(n) divides (n -1) iff 'n' is prime.
                  > > > > > > (proof, backward)
                  > > > > > > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
                  > > > > > > divides (n -1), exactly.
                  > > > > > > (proof, forward)
                  > > > > > > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
                  > > > > > > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
                  > > > > > > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
                  > > > > > > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
                  > > > > > > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
                  > > > > > > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
                  > > > > > > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
                  > > > > > > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
                  > > > > > > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
                  > > > > > > cannot be more co-prime divisors of C than there are the total number
                  > > > > > > of possible divisors! thus, by contradiction, 'n' must be prime.
                  > > > > > > ...
                  > > > > > > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
                  > > > > > > marker on the back of a receipt for car repairs. at first glance, it
                  > > > > > > looks as though the problem needs to be solved using Fermat's little
                  > > > > > > theorem; that is DEFINITELY NOT the case. enjoy!
                  > > > > > > ...
                  > > > > > > *QED
                  > > > > > > 05/21/2013
                  > > > > > > ...
                  > > > > > > Bill Bouris
                  > > > > > > www.oddperfectnumbers.com
                  > > > > > >
                  > > > > > >
                  > > > > > >
                  > > > > >
                  > > > > >
                  > > > > >
                  > > > > > --
                  > > > > > La verdad (blog de raciocinio político e información
                  > > > > > social)<http://josebrox.blogspot.com/>
                  > > > > >
                  > > > > >
                  > > > > > [Non-text portions of this message have been removed]
                  > > > > >
                  > > > >
                  > > >
                  > >
                  >
                • John
                  The readability is considerably improved apart from some Phis which have sneaked in. ... You say ... a step which I cannot follow. it is also evident for the
                  Message 8 of 8 , May 26 6:29 PM
                  • 0 Attachment
                    The readability is considerably improved apart from some Phis which have sneaked in.

                    --- In primenumbers@yahoogroups.com, "John" <mistermac39@...> wrote:
                    >
                    > For the sake of readability, I have set out the "backwards" part of the proof.

                    You say

                    > Ø implies that k= b^(j-1)= 1.

                    a step which I cannot follow. it is also evident for the preceding part of the proof that your k cannot take the value zero, otherwise we have some illegal division. And how, k could take any negative value, I cannot see. Now, k can take the value 1 only in the case of n being prime. Otherwise, the very meaning of Euler's Totients is debauched.

                    So, having by reasoning which eludes me, having shown that k can only have the value 1, which if true, of itself is sufficient to show that n cannot be composite without needing further comment.

                    Further, not only do you conclude that k = 1, you also conclude that b = C, and somehow, j completely fades from view, unless you are saying that j is also equal to 1, a conclusion that Lehmer himself proved by showing that n must be square free.

                    Now, if b = C, n is not square free. How you reasoned that b = C from the preceding part of the proof also eludes me. But, if your reasoning is correct, you have indeed come to the same conclusion as Lehmer. But, you have gone even one better, you have also proven that phi(b) = b + 1, which indeed is a contradiction.

                    So, the only thing missing is my failure to see that the conclusion that

                    > Ø implies that k= b^(j-1)= 1

                    part of the proof.

                    Any requirement that it is not necessary, as some say, to dismiss the possibility that n may be a Carmichael Number, (which would imply that C needs to be a composite number)is thus rendered spurious as far as I can see, as essentially you are using "reductio ad adsurdum" as your method of proof.

                    Thus ends my exercise in senile dementia aversion, which at the age of 73, I wish to try and avoid.


                    --- In primenumbers@yahoogroups.com, "John" <mistermac39@...> wrote:
                    >
                    > For the sake of readability, I have set out the "backwards" part of the proof.
                    >
                    > let phi(n) divide (n -1), such that k*phi(n)= (n -1), and
                    > Ø assume 'n' is composite,
                    > Ø n= b^j*C where gcd(b^j, C)= 1
                    > Ø and
                    > Ø k <> 1.
                    > Ø
                    > Ø so, k*phi(b^j*C) = b^j*C -1,
                    > Ø k*phi(b^j)*phi(C)= b^j*C -1,
                    > Ø (k*b^j -k*b^(j-1))*phi(C)= b^j*C -1,
                    > Ø or... phi(C)=[b^j*C -1] / [k*b^j - k*b^(j-1)]
                    > Ø
                    > Ø implies that k= b^(j-1)= 1.
                    > Ø
                    > Ø > hence, phi(C) = (bC -1) / (b -1),
                    > Ø b= C,
                    > Ø and phi(C)= C +1.
                    > Ø
                    > but, 'C' cannot have more co-prime divisors than it has a
                    > > > total number of /*possible*/ divisors. thus, 'n' must be
                    > > > prime.enjoy!
                    >
                    >
                    > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                    > >
                    > > i forgot to add the word 'possible' divisors.
                    > >
                    > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                    > > >
                    > > > the third time's a charm... I had to re-write the proof.
                    > > > now, it truly IS a classic. it's done, done, done!
                    > > > ...
                    > > > D. H. Lehmer's (1932) Totient Conjecture Stated: phi(n)
                    > > > divides (n -1) iff 'n' is prime.
                    > > > (proof, backward)
                    > > > if 'n' is prime, then phi(n) = (n -1) by definition, and
                    > > > phi(n) | (n -1), exactly as desired.
                    > > > (proof, forward)
                    > > > let phi(n) divide (n -1), such that k*phi(n)= (n -1), and
                    > > > assume 'n' is composite, n= b^j*C where gcd(b^j, C)= 1 and
                    > > > k <> 1. so, k*phi(b^j*C) = b^j*C -1, k*phi(b^j)*phi(C)= b^j
                    > > > *C -1, (k*b^j -k*b^(j-1))*phi(C)= b^j*C -1, or... phi(C)=
                    > > > [b^j*C -1] / [k*b^j - k*b^(j-1)] implies that k= b^(j-1)= 1.
                    > > > hence, phi(C) = (bC -1) / (b -1), b= C, and phi(C)= C +1.
                    > > > but, 'C' cannot have more co-prime divisors than it has a
                    > > > total number of /*possible*/ divisors. thus, 'n' must be
                    > > > prime.enjoy!
                    > > > ...
                    > > > *QED
                    > > > 05/26/2013
                    > > > ...
                    > > >
                    > > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                    > > > >
                    > > > > ...
                    > > > > D. H. Lehmer's (1932) Totient Conjecture Stated:
                    > > > > phi(n) divides (n -1) iff 'n' is prime.
                    > > > > (proof, backward)
                    > > > > if 'n' is prime, then phi(n) = (n -1) by definition, and
                    > > > > phi(n) | (n -1), exactly as desired.
                    > > > > (proof, forward)
                    > > > > let phi(n) divide (n -1), or k*phi(n)= (n -1), and assume
                    > > > > 'n' is composite, or n= b^j*C where gcd(b^j, C)= 1 and k <> 1.
                    > > > > so, k*phi(b^j*C) implies k*b^j*C -k*b^(j-1)*C -1= b^j*C -1 and
                    > > > > adding one to... and dividing by b^(j-1) on... both sides, and
                    > > > > rearranging terms, we have... k= bC / [bC -1]. therefore, k= 1,
                    > > > > and phi(b^j*C) = b^j*C -1, and b^j*C -b^(j-1)*C = b^j*C -1 im-
                    > > > > plies b^(j-1)*C = 1, or b= C= 1, and 'n' isn't composite. thus,
                    > > > > 'n' must be prime. enjoy!
                    > > > > ...
                    > > > > *QED
                    > > > > 05/25/2013
                    > > > > ...
                    > > > > thanks, Jose B.; it's done! rewards, Bill
                    > > > >
                    > > > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                    > > > > >
                    > > > > > here's what the missing portion would be similar to... let
                    > > > > > k <> 1, n= b^j*C is composite, and k*phi(b^j*C)= b^j*C -1
                    > > > > > where gcd(b, C)= 1. then we'd find that phi(C)= [b^j*C -1]
                    > > > > > / [k*b^(j-1)*(b- 1) -1]... forcing us to realize that k= 1
                    > > > > > and phi((b-1)/b) = phi(C) which is still a contradiction!
                    > > > > > we don't have to doubt the eventual answer.(5 more minutes)
                    > > > > > www.oddperfectnumbers.com
                    > > > > >
                    > > > > > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
                    > > > > > >
                    > > > > > > I've presented to you the basics of how the proof would
                    > > > > > > take place. If you'd like to work through a few details
                    > > > > > > to improve upon my idea, then do so. If you aren't part
                    > > > > > > of the solution, then you're part of the problem...
                    > > > > > > rewards, Bill
                    > > > > > >
                    > > > > > > --- In primenumbers@yahoogroups.com, Jose Ramón Brox <ambroxius@> wrote:
                    > > > > > > >
                    > > > > > > > Note that phi(bC)=phi(b)phi(C) if and only if gcd(b,C)=1, but it could be
                    > > > > > > > that C=bC'.
                    > > > > > > >
                    > > > > > > > Regards,
                    > > > > > > > Jose Brox
                    > > > > > > >
                    > > > > > > >
                    > > > > > > > 2013/5/24 leavemsg1 <leavemsg1@>
                    > > > > > > >
                    > > > > > > > > **
                    > > > > > > > >
                    > > > > > > > >
                    > > > > > > > > a copy of this proof can be found at www.oddperfectnumbers.com
                    > > > > > > > > ...
                    > > > > > > > > D. H. Lehmer's (1932) Totient Conjecture Stated:
                    > > > > > > > > phi(n) divides (n -1) iff 'n' is prime.
                    > > > > > > > > (proof, backward)
                    > > > > > > > > if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n)
                    > > > > > > > > divides (n -1), exactly.
                    > > > > > > > > (proof, forward)
                    > > > > > > > > let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime;
                    > > > > > > > > let n= bC where 'b' is one prime factor of 'n', and 'C' is one or more
                    > > > > > > > > other factors, combined, and k = 1. assume that k <> 1; so, k*phi(bC)
                    > > > > > > > > = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1]
                    > > > > > > > > / [b*k -k]; it has to be true that k= C= 1 in order for phi(C) to not
                    > > > > > > > > be fractional while phi(1) = (b -1) / (b -1) = 1 is verified. hence,
                    > > > > > > > > 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1]
                    > > > > > > > > / [b -1] implies that C= b as the only solution. phi(C) = [b^2 -1] /
                    > > > > > > > > [b -1]; therefore, it's not possible to have phi(C)= C +1, (OR) there
                    > > > > > > > > cannot be more co-prime divisors of C than there are the total number
                    > > > > > > > > of possible divisors! thus, by contradiction, 'n' must be prime.
                    > > > > > > > > ...
                    > > > > > > > > it took me 20 minutes to solve Lehmer's Conjecture using a sharpie
                    > > > > > > > > marker on the back of a receipt for car repairs. at first glance, it
                    > > > > > > > > looks as though the problem needs to be solved using Fermat's little
                    > > > > > > > > theorem; that is DEFINITELY NOT the case. enjoy!
                    > > > > > > > > ...
                    > > > > > > > > *QED
                    > > > > > > > > 05/21/2013
                    > > > > > > > > ...
                    > > > > > > > > Bill Bouris
                    > > > > > > > > www.oddperfectnumbers.com
                    > > > > > > > >
                    > > > > > > > >
                    > > > > > > > >
                    > > > > > > >
                    > > > > > > >
                    > > > > > > >
                    > > > > > > > --
                    > > > > > > > La verdad (blog de raciocinio político e información
                    > > > > > > > social)<http://josebrox.blogspot.com/>
                    > > > > > > >
                    > > > > > > >
                    > > > > > > > [Non-text portions of this message have been removed]
                    > > > > > > >
                    > > > > > >
                    > > > > >
                    > > > >
                    > > >
                    > >
                    >
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