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Re: [PrimeNumbers] odd proof

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  • Jack Brennen
    Peter: I haven t seen your proof at all, but Bill did share his with me separately, and if his is anything like yours (claiming that a particular equation has
    Message 1 of 9 , Oct 22, 2009
      Peter:

      I haven't seen your proof at all, but Bill did share his with me separately,
      and if his is anything like yours (claiming that a particular equation has
      no solutions), I would just ask that you make sure that your proof explains
      why this is solvable:

      N = p^2*q^2*r^2*s^2*X p,q,r,s prime, X composite
      2N = (p^2+p+1)*(q^2+q+1)*(r^2+r+1)*(s^2+s+1)*(X+1)

      but this is not:

      N = p^2*q^2*r^2*s^2*t p,q,r,s,t prime
      2N = (p^2+p+1)*(q^2+q+1)*(r^2+r+1)*(s^2+s+1)*(t+1)


      The only difference being that the composite X in the first example
      is replaced by the prime t in the second. Or, in other words, the
      proof better make use of the fact that the prime factors are
      irreducible.
    • Bill Bouris
      Dear Mr. Brennen, if I thought that the factors for each f1, f2, etc. were all prime, even for a second, then I wouldn t have considered the notion of proof
      Message 2 of 9 , Oct 23, 2009
        Dear Mr. Brennen, if I thought that the factors for each f1, f2, etc.
        were all prime, even for a second, then I wouldn't have considered the
        notion of proof using anyone's or Mr. Lesala's definition... look at
        the 'even' perfect number for eg.... 1+2+4+7+14 = 28; I DO know that at
        least a few of the factors must be composite... so you are way off base
        and probably cannot find a counter-example using my correction to the
        definition; yes as it turns out... letting q= 2r +1 where r>0 is natural,
        then 3 cannot be one of the first few factors.
        what's the matter... can't you find a counter-example; I don't make the rules... and they are like baseball; three strikes, and I'm out; so far,
        you have one strike against me... you were able to find at least one
        counter-example when q is one, but then 'r' would have to be non-exist-
        ant. Oh... it includes all the factors, you just don't like the defin-
        tion; I desparately want to remain humble in my victory and brave in my
        defeat; here's another baseball analogy that often comes into play...
        you win some, you lose some, and some get rained out. I think that with-
        out your next counter-example, your counter-logic puts you into the 2nd
        category ??? or if you think that all factors have to be primes, then
        your counter-logic puts you in the last category; we still have a game...
        you NEED to find your counter-example when you fully understand the
        technique; thanks for even listening to me; please, please, follow the
        definition.


        --- On Thu, 10/22/09, Jack Brennen <jfb@...> wrote:

        > From: Jack Brennen <jfb@...>
        > Subject: Re: odd proof
        > To: "Bill Bouris" <leavemsg1@...>
        > Date: Thursday, October 22, 2009, 10:23 AM
        > You seemed to miss the point with
        > both of them...
        >
        > 1 is a factor of every natural number, and is in the factor
        > list of
        > every natural number.  And in your own proof
        > methodology, you
        > explicitly start the "f" sequence at the number AFTER the
        > 1, so
        > in neither of the cases which I gave would q be equal to
        > 1.  Instead
        > q would be equal to 3.
        >
        > The point behind both was that your proof doesn't seem to
        > use the
        > very important fact that the sum 1 + f1 + ... + fn = N must
        > include
        > ALL of the factors of N, not just some of them. 
        > Because I've given
        > you two examples where the sum works over a subset of the
        > factors
        > of N, even under the requirement that f1*fn = f2*f(n-1) =
        > ... = N.
        > So unless your proof actually *uses* the idea that the odd
        > perfect
        > number's sum must include *all* of the factors, then it's
        > flawed.
        >
        >
        >
        > Bill Bouris wrote:
        > > I think that the answer for your counter-
        > counter-example is that...
        > > if I let q = 2r+1 where r >0, then it would avoid
        > letting you choose
        > > that q = 1; r just has to be a natural number. 
        > you are very smart;
        > > did you ever think of changing your name to Johnny
        > Quest or Race Banner?
        > > j/k  I think that if r is qualified as a natural
        > number, then those two
        > > cases you mentioned will not prevail.
        > >
        > > --- On Wed, 10/21/09, Jack Brennen <jfb@...>
        > wrote:
        > >
        > >> From: Jack Brennen <jfb@...>
        > >> Subject: Re: odd proof
        > >> To: "Bill Bouris" <leavemsg1@...>
        > >> Date: Wednesday, October 21, 2009, 6:52 PM
        > >> Or the much shorter example:
        > >>
        > >> N=12285
        > >>
        > >> 1
        > >> 3 * 4095 == 12285
        > >> 5 * 2457 == 12285
        > >> 7 * 1755 == 12285
        > >> 9 * 1365 == 12285
        > >> 13 * 945 == 12285
        > >> 21 * 585 == 12285
        > >> 45 * 273 == 12285
        > >> 63 * 195 == 12285
        > >> 91 * 135 == 12285
        > >> 105 * 117 == 12285
        > >>
        > >> And:
        > >>
        > >>
        > 1+3+5+7+9+13+21+45+63+91+105+117+135+195+273+585+945+1365+1755+2457+4095
        > >> is equal to 12285
        > >>
        > >> Now obviously neither 12285 nor 198585576189 is an
        > odd
        > >> perfect number,
        > >> but both of them seem to have an expression in a
        > sum format
        > >> that your
        > >> proof claims is impossible?
        > >>
        > >>
        > >>
        > >> Jack Brennen wrote:
        > >>> Or to rephrase what I was trying to say a
        > little
        > >> better...
        > >>> Your proof apparently is trying to prove that
        > it's
        > >> impossible to have
        > >>> N = 1 + f1 + f2 + ... + fj + (N/fj) + ... +
        > (N/f2) +
        > >> (N/f1)
        > >>> where N is odd, and where the first j terms of
        > the
        > >> sequence
        > >>> (after the 1) are divisors of N < sqrt(N),
        > and the
        > >> last j terms
        > >>> of the sequence are respectively equal to N
        > divided by
        > >> the
        > >>> "mirrored" values at the front.
        > >>>
        > >>> But the number N=198585576189 has a set of
        > 161
        > >> divisors which meet the
        > >>> above requirements and which sum up to N:
        > >>>
        > >>> [1, 3, 7, 9, 11, 13, 21, 33, 39, 49, 63, 77,
        > 91, 99,
        > >> 117, 121, 143, 147,
        > >>> 169, 231, 273, 363, 429, 441, 507, 539, 637,
        > 693, 819,
        > >> 847, 1001, 1089,
        > >>> 1183, 1287, 1521, 1573, 1617, 1859, 1911,
        > 2541, 3003,
        > >> 3549, 4719, 4851,
        > >>> 5577, 5733, 5929, 7007, 7623, 8281, 9009,
        > 10647,
        > >> 11011, 13013, 14157,
        > >>> 16731, 17787, 20449, 21021, 22021, 24843,
        > 33033,
        > >> 39039, 53361, 61347,
        > >>> 63063, 66063, 74529, 77077, 91091, 99099,
        > 117117,
        > >> 143143, 154147,
        > >>> 184041, 198189, 231231, 242231, 273273,
        > 286273,
        > >> 429429, 462441, 693693,
        > >>> 726693, 819819, 858819, 1002001, 1079029,
        > 1288287,
        > >> 1387323, 1695617,
        > >>> 2003911, 2180079, 2576457, 2664541, 3006003,
        > 3149003,
        > >> 3237087, 3721549,
        > >>> 5086851, 6011733, 7993623, 9018009, 9447009,
        > 9711261,
        > >> 11164647,
        > >>> 11869319, 14027377, 15260553, 18035199,
        > 18651787,
        > >> 22043021, 23980869,
        > >>> 26050843, 28341027, 33493941, 34639033,
        > 35607957,
        > >> 40937039, 42082131,
        > >>> 55955361, 66129063, 78152529, 103917099,
        > 106823871,
        > >> 122811117,
        > >>> 126246393, 130562509, 154301147, 167866083,
        > 182355901,
        > >> 198387189,
        > >>> 234457587, 242473231, 286559273, 311751297,
        > 368433351,
        > >> 391687527,
        > >>> 450307429, 462903441, 547067703, 727419693,
        > 859677819,
        > >> 1175062581,
        > >>> 1350922287, 1388710323, 1641203109,
        > 1697312617,
        > >> 2005914911, 2182259079,
        > >>> 2579033457, 3152152003, 4052766861,
        > 5091937851,
        > >> 6017744733, 9456456009,
        > >>> 15275813553, 18053234199, 22065064021,
        > 28369368027,
        > >> 66195192063]
        > >>> Jack Brennen wrote:
        > >>>> Your proof doesn't seem to use the prime
        > >> factorization at all,
        > >>>> which is obviously a problem. 
        > Consider how
        > >> your proof will show that
        > >>>> the number 198585576189 is not perfect.
        > >> Because it "almost" is
        > >>>> perfect -- if you consider the number
        > 22021 to be
        > >> prime.  It has a
        > >>>> set of 161 proper divisors which meet all
        > of the
        > >> pairwise requirements
        > >>>> for being an odd perfect number.  The
        > only
        > >> reason it's not perfect is
        > >>>> because 22021 has divisors.  If you
        > "pretend"
        > >> that 22021 is prime,
        > >>>> you have an odd perfect number.  :)
        > >> Your proof would have to have
        > >>>> some sort of clause which disqualifies
        > this number
        > >> from being perfect,
        > >>>> not because the factors don't add up
        > (because they
        > >> do!), but because
        > >>>> one of the "prime" factors isn't actually
        > >> prime...
        > >>>>    Jack
        > >>>>
        > >>>>
        > >>>> Bill Bouris wrote:
        > >>>>> Dear Mr. Brennen,
        > >>>>>
        > >>>>> I’m glad that someone is genuinely
        > >> interested in what I have to say...
        > >>>>> here’s a rather funny analogy which
        > >> describes my involvement in the OPN
        > >>>>> don't exist proof:
        > >>>>>
        > >>>>> the director invites me to a preview
        > of his
        > >> movie for which I am the
        > >>>>> special effects person; we sit down to
        > watch
        > >> it, and I give him a ton of
        > >>>>> encouragement to produce the movie and
        > then
        > >> show him how to capture the
        > >>>>> crucial moments of a particular scene
        > that
        > >> will make the production a
        > >>>>> smash hit!
        > >>>>>
        > >>>>> he then gets scared with the prospect
        > of
        > >> having to share the credits of
        > >>>>> the movie with a person whom he
        > barely
        > >> knows... so, he makes a mad dash
        > >>>>> for the exit, but not until he has
        > heard my
        > >> idea for capturing that most
        > >>>>> important scene.
        > >>>>>
        > >>>>> now, I’m just trying to get him to
        > >> acknowledge my contribution before he
        > >>>>> runs off to the box office... listing
        > himself
        > >> as the sole writer, (pro-ducer), and director of
        > the movie.
        > >>>>> personal greed is interfering with the
        > making
        > >> of a great movie and every-
        > >>>>> one receiving due credit.
        > >>>>>
        > >>>>> end of story... neither of the people
        > in this
        > >> dilemma are as important as
        > >>>>> the movie itself.
        > >>>>>
        > >>>>> I really do believe that no one can
        > fully
        > >> understand anything about
        > >>>>> something all by themselves.
        > >>>>>
        > >>>>> I’ve sought out other resources for
        > >> publishing the one-page proof and have
        > >>>>> listed both of Peter's and my names on
        > the
        > >> copies I have given to them and
        > >>>>> to you; spread it around... the source
        > that I
        > >> told him to submit to will
        > >>>>> not fully endorse his proof if it's
        > submitted
        > >> to a large number of people;
        > >>>>> there submition policy 'clearly'
        > stated that
        > >> fact.
        > >>>>> Bill Bouris; pasted below... and
        > attached
        > >> also.
        > >>>>>
        > >>
        > 012345678901234567890123456789012345678901234567890123456789012345
        > >>>>> hypothesis: suppose that odd perfect
        > numbers
        > >> do exist...
        > >>>>> both you and I MUST define them as
        > the
        > >> following:
        > >>>>> let an odd perfect number N = 1 + f1 +
        > f2 +
        > >> ... + f(n-1) + fn
        > >>>>> where the fi's are odd factors of N
        > with the
        > >> constraint f1*fn=
        > >>>>> f2*f(n-1)= ... = N, and f1= q + k1,
        > f2= q +
        > >> k2, ..., f(n-1)= q
        > >>>>> + k(n-1), fn= q + kn where q is some
        > odd
        > >> natural number, and
        > >>>>> all ki's are 'even' natural numbers,
        > and i
        > >> ranges from 1 to some
        > >>>>> 'even' natural number n; (extremely
        > well
        > >> defined! no mistakes.)
        > >>>>> the simplest case... although most
        > math
        > >> persons would agree that
        > >>>>> it doesn't exist from a cursory
        > glance... I'm
        > >> proving it!
        > >>>>> if N = 1 + f1 + f2 and N = f1*f2, then
        > someone
        > >> will notice that
        > >>>>> the discriminant of their quadratic
        > solution
        > >> does not have an in-
        > >>>>> tegral value; so, it's quite
        > elementary, for
        > >> sure.
        > >>>>> fact: 'n' has to be 'even', because
        > adding an
        > >> odd number of odd
        > >>>>> factors together plus one would result
        > in 'N'
        > >> being 'even'.
        > >>>>> if n >= 4, then all generic
        > pairings
        > >> f(i-3)*f(i-2)= f(i-1)*fi
        > >>>>> would result in the following:
        > >>>>>
        > >>>>> q = (k(i-3)*k(i-2)-k(i-1)*ki)/((ki
        > +k(i-1)
        > >> -k(i-2) -k(i-3)) sim-
        > >>>>> plifying into... q = (a*b-c*d)/(d +c
        > -b -a)
        > >> where a>0, b>0, c>0,
        > >>>>> d>0 each represent some 'even' ki's
        > from
        > >> our definition.
        > >>>>> 'q' cannot be odd; letting q= 2r +1 =
        > >> (a*b-c*d)/(d +c -b -a) would
        > >>>>> imply that (a*b-c*d) = (2r+1)*d
        > +(2r+1)*c
        > >> -(2r+1)*b - (2r+1)*a  or
        > >>>>> that at least one ki's must be odd for
        > 'q' to
        > >> be odd; it doesn't
        > >>>>> equate as long as all ki's are
        > 'even'.
        > >>>>>
        > >>>>> conclusion: odd perfect numbers
        > cannot
        > >> exist...
        > >>
        > 012345678901234567890123456789012345678901234567890123456789012345
        > >>>>> *QED
        > >>>>>
        > >>>>> Best regards,
        > >>>>> Peter Lesala; Lesotho, Africa
        > >>>>> Bill Bouris; Aurora, IL
        > >>>>> both, co-contributors  10/21/09
        > >>>>>
        > >>>>>
        > >>>>>       
        > >>>>
        > >>>
        > >>
        > >
        > >
        > >       
        > >
        > >
        >
        >
      • djbroadhurst
        ... In mathematics you are out first strike. Also in cricket: http://www.gap-system.org/~history/Extras/Hardy_cricket_teams.html David
        Message 3 of 9 , Oct 23, 2009
          --- In primenumbers@yahoogroups.com,
          Bill Bouris <leavemsg1@...> wrote:

          > three strikes, and I'm out

          In mathematics you are out first strike. Also in cricket:
          http://www.gap-system.org/~history/Extras/Hardy_cricket_teams.html


          David
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