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• 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.
• 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-
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
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
> 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:
> 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
> >>>>> 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
> >> 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
> >>>>>
> >>>>>
> >>>>>
> >>>>
> >>>
> >>
> >
> >
> >
> >
> >
>
>
• ... 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