PING James Harris
Arturo Magidin
James Harris <jst...@msn.com> wrote:
>If you or anyone else can find a factorization of
>
> 5g^6 - 144 g^4 + 312 g^2 - 169
>
>such that
>
> 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
>
>where the c's and d's are algebraic integers, then not only am I wrong
>in disputing your claim, but I don't have proof of Fermat's Last
>Theorem.
I still say you are completely wrong about how the factorization of
this polynomial relates to your "proof." The existence or lack-thereof
of a factorization as above is not material to your errors.
But anyway. I don't have a full answer yet, so let me just post what I
have. Maybe some of the folks at sci.math.symbolic can give me the
final points I'm missing.
It is enough to factor the polynomial
f(x) = 5X^3 - 144X^2 + 312 X - 169
into linear terms with algebraic integer coefficients. One
factorization into linear terms (not with algebraic integer
coefficients) is:
f(x) = 5 (x-(r1/5))(x-(r2/5))(x-(r3/5))
where r1, r2, r3 are the roots of
g(y) = y^3 - 144 y^2 + 1560 y - 4225.
To verify this, note that if r1, r2, r3 are the roots given, then
r1*r2*r3 = 4225
r1*r2 + r1*r3 + r2*r3 = 1560
r1+r2+r3 = 144
so 5*(r1/5)(r2/5)(r3/5) = 4225/25 = 169
5*( (r1/5)(r2/5) + (r1/5)(r3/5) + (r2/5)(r3/5)) = 1560/5 = 312
5*(r1/5) + (r2/5) + (r3/5) = 144,
giving the right factorization.
So let r be a root of g(y). Let K=Q(r) be the number field generated
by r, which is an algebraic integer. Let R be the ring of integers of
Q(r).
I need the gcd of r and 5 in the ring of all algebraic integers. This
can be obtained by letting h be the class number of R, and letting d
be a generator of the principal ideal (r,5)^h. Then d^{1/h} is the
gcd. If someone can provide me with either h, d, both, or simply an
n>0 and a d in R such that (r,5)^n=(d), that should settle the
problem. d should be expressible as a polynomial of degree at most 2
in r.
This is a fairly annoying process to do by hand, so I'm wondering if
someone out there can use a symbolic algebra package to figure it out.
I will try, as time permits, to figure it out myself.
======================================================================
"If Yeats was right when he wrote 'The best lack all conviction, while
the worst are full of passionate intensity,' then investigations by
Boris Fischoff and others on overconfidence suggest that most of us
aren't very admirable. [...] We're so often cocksure of our decisions,
actions, and beliefs because we fail to look for counterexamples, pay
no attention to alternative views and their consequences, distort our
memories and the evidence, and are seduced by our own explanatory
schemes."
-- John Allen Paulos, _Once Upon a Number: The Hidden Mathematical
Logic of Stories_
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Greg Doyle
proven. Prove the Riemann hypothesis or the Goldbach Conjecture or
something.
GREG
"James Harris" <jst...@msn.com> wrote in message
news:3c65f87.03042...@posting.google.com...
> mag...@math.berkeley.edu (Arturo Magidin) wrote in message
news:<b80vcp$2uh5$1...@agate.berkeley.edu>...
> > In article <3c65f87.03042...@posting.google.com>,
> > James Harris <jst...@msn.com> wrote:
> >
> > [.James proposes a polynomial.]
> >
> > > 5 g^6 - 144 g^4 + 2426 g^2 - 169
>
> As noted by another poster that should be
>
> 5g^6 - 144 g^4 + 312 g^2 - 169.
> >
> > I haven't tried factoring this polynomial. But before I spend my time
> > doing so, let me ask you:
> >
> > Suppose (imagine, fantasize if you will) that someone succeeds in
> > exhibiting a factorization of this polynomial into linear terms, each
> > with algebraic integer coefficients.
> >
> > Will you then (a) admit your argument against the theorem is incorrect
> > ->somewhere<-, even if you can't figure out where? Or (b) dismiss it
> > and try some other example?
>
> <deleted>
>
> If you or anyone else can find a factorization of
>
> 5g^6 - 144 g^4 + 312 g^2 - 169
>
> such that
>
> 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
>
> where the c's and d's are algebraic integers, then not only am I wrong
> in disputing your claim, but I don't have proof of Fermat's Last
> Theorem.
>
> If ANYONE can produce the c's and d's such that they are algebraic
> integers, then that's it.
>
> I would *immediately* not only withdraw my claim of having proven
> Fermat's Last Theorem, but would also pull what I have on the subject,
> including my object pages from the web.
>
> Further, I would apologize to you Magidin in a POST, for questioning
> your honesty and mathematical ability.
>
> Is that good enough for you?
>
> However, note that I am NOT wrong, and you are. And I'm tired of you
> wasting my time with your interminable debating and posting!!!
>
>
> James Harris
Dot
> If you or anyone else can find a factorization of
>
> 5g^6 - 144 g^4 + 312 g^2 - 169
>
> such that
>
> 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
>
> where the c's and d's are algebraic integers, then not only am I wrong
> in disputing your claim, but I don't have proof of Fermat's Last
> Theorem.
OK, I'll give such a factorization. It seems that the easiest way
to give the coefficients so that somone without specialized software
can verify the example is to specify them as complex numbers to high
precision and specify their minimal polynomials. See below.
> If ANYONE can produce the c's and d's such that they are algebraic
> integers, then that's it.
>
> I would *immediately* not only withdraw my claim of having proven
> Fermat's Last Theorem, but would also pull what I have on the subject,
> including my object pages from the web.
>
> Further, I would apologize to you Magidin in a POST, for questioning
> your honesty and mathematical ability.
>
> Is that good enough for you?
I don't know whether that would be good enough for Arturo, but it would
certainly be a high point of my day to see such posts. But I'm not
holding my breath, because I expect that James will find *something* to
argue about...
The computations below are set up for the computer algebra system Magma,
but the numbers can be plugged into your favorite program. Or heck, use
your calculator. All of the coefficients happen to live in a totally
real field, so we don't even need to use complex numbers, just real
numbers.
// Note that n1 is negative, and all the other values are
// positive. I chose things this was to make the comment
// at the end of the post work out.
n1 := -0.0331807983045615453788926498722140138507770879;
n2 := 8.6290051111993043659968319502570646814892496758;
n3 := 45.4041756871052571793820606996151493323615273716;
d1 := 0.0346096699784644515265124458385748896347812760;
d2 := 1.6764738885316136065575056637043273138144707330;
d3 := 38.5381400364861364772811079276200700344818784657;
// Are these numbers algebraic integers?
// Note that the following monic polynomials evaluate to 0
// to at least 34 decimal places.
n1^6 - 2136*n1^4 + 153504*n1^2 - 169;
n2^6 - 2136*n2^4 + 153504*n2^2 - 169;
n3^6 - 2136*n3^4 + 153504*n3^2 - 169;
d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
d2^6 - 1488*d2^4 + 4176*d2^2 - 5;
d3^6 - 1488*d3^4 + 4176*d3^2 - 5;
R<x>:=PolynomialRing(RealField());
// Have we factored the desired polynomial? Multiply out the
// following expression:
(d1*x - n1)*(d1*x + n1)*(d2*x - n2)*(d2*x + n2)*(d3*x - n3)*(d3*x + n3);
// To 40 decimal places, the result is
// 5*x^6 - 144*x^4 + 312*x^2 - 169
// A note for those who want to check things algebraically,
// and not just numerically:
// The n's are related to the d's via
// 10683*n = -22*d^4 + 33001*d^2 - 394
// which leads to
// 10683*n^2 = -181*d^4 + 283648*d^2 - 328
Dot
wrote:
> n1 := -0.0331807983045615453788926498722140138507770879;
> n2 := 8.6290051111993043659968319502570646814892496758;
> n3 := 45.4041756871052571793820606996151493323615273716;
> [...]
> // Are these numbers algebraic integers?
> // Note that the following monic polynomials evaluate to 0
> // to at least 34 decimal places.
>
>
> n1^6 - 2136*n1^4 + 153504*n1^2 - 169;
> n2^6 - 2136*n2^4 + 153504*n2^2 - 169;
> n3^6 - 2136*n3^4 + 153504*n3^2 - 169;
I didn't check those polynomials for irreducibility. Even better, the
following monic polynomials evaluate to 0 to at least 40 decimal
places:
n1^3 - 54*n1^2 + 390*n1 + 13;
n2^3 - 54*n2^2 + 390*n2 + 13;
n3^3 - 54*n3^2 + 390*n3 + 13;
-- Dot.
Dot
> Dot <d...@at.dot.com> wrote in message
> news:<210420031722359058%d...@at.dot.com>...
> > In article <3c65f87.03042...@posting.google.com>, James
> > Harris <jst...@msn.com> wrote:
> >
> > > If you or anyone else can find a factorization of
> > >
> > > 5g^6 - 144 g^4 + 312 g^2 - 169
> > >
> > > such that
> > >
> > > 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
> > >
> > > where the c's and d's are algebraic integers, then not only am I wrong
> > > in disputing your claim, but I don't have proof of Fermat's Last
> > > Theorem.
> >
> > OK, I'll give such a factorization. It seems that the easiest way
> > to give the coefficients so that somone without specialized software
> > can verify the example is to specify them as complex numbers to high
> > precision and specify their minimal polynomials. See below.
>
> [snip]
>
> Decimals don't do it for me.
>
> [snip]
>
> Ok, I'll consider the possibility that you've found what I still say
> doesn't exist, but I need EXACT solutions.
>
> [snip]
>
> Use *much* detail please.
>
> [snip]
>
> Ok, now here it looks interesting to me. Why don't you elaborate?
I'm happy to elaborate, James. I'll spell it out for you in complete
detail, with exact numbers. To verify that what I write is correct,
all you need to be able to do is multiply polynomials together. You
might complain that that is too much work to expect of you, so I have
included Magma code to verify the work. You might complain that you
don't have Magma, and that it costs money to obtain it from
http://magma.maths.usyd.edu.au/magma/ , so I have included Pari code to
verify the work. You might complain that you don't have Pari. In that
case, you can go to http://www.parigp-home.de/ and download a copy for
free.
I await your apology to Arturo. But I'm still not holding my breath.
-- Dot.
We will work in the number field defined by the polynomial
y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
+ 28575240325442631201*y^4 - 413298187541059968292500*y^2
+ 2460349402142038081821721600;
We will let alpha be a root of this polynomial, and we will define
algebraic integers d1, d2, d3, n1, n2, n3 in terms of alpha.
(Taking alpha to be the real root that is approximately
154.451496990997802982236692688237765859998280922
gives the numerical examples I gave earlier.
MAGMA CODE:
R<y>:=PolynomialRing(Rationals());
f := y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
+ 28575240325442631201*y^4 - 413298187541059968292500*y^2
+ 2460349402142038081821721600;
K<alpha>:=NumberField(f);
// Define our coefficients:
d1 := (1/1401502976449936390824092894020468134024616320)*
(- 116858525730451586855252258697*alpha^11
+ 21546461434737374316017502217792*alpha^10
+ 22109717335034240282753967154849470*alpha^9
- 4076616001448173117175959963381396768*alpha^8
- 1647209069948549841509121657588818730783*alpha^7
+ 303715513906432298157776614831268988666880*alpha^6
+ 60404344550512455043182143206734511673314206*alpha^5
- 11137524588944779726233177935687516957256627968*alpha^4
- 1089475637849489503980551259022753852684824408489*alpha^3
+ 200881897616666532603687714008711443626111508690592*alpha^2
+ 7719396850213757345324975434862758616298359729245140*alpha
- 1423343550253117792241514076181168733734615795436123200);
d2 := (1/1401502976449936390824092894020468134024616320)*
(- 116858525730451586855252258697*alpha^11
- 21546461434737374316017502217792*alpha^10
+ 22109717335034240282753967154849470*alpha^9
+ 4076616001448173117175959963381396768*alpha^8
- 1647209069948549841509121657588818730783*alpha^7
- 303715513906432298157776614831268988666880*alpha^6
+ 60404344550512455043182143206734511673314206*alpha^5
+ 11137524588944779726233177935687516957256627968*alpha^4
- 1089475637849489503980551259022753852684824408489*alpha^3
- 200881897616666532603687714008711443626111508690592*alpha^2
+ 7719396850213757345324975434862758616298359729245140*alpha
+ 1423343550253117792241514076181168733734615795436123200);
d3 := (1/65595009662544996294303701863730606291520)*
(- 240064724443796467*alpha^11
+ 55326448296076857699530*alpha^9
- 5309131403735081126863921733*alpha^7
+ 274489395789430590995126949847466*alpha^5
- 8357164018995647094694127739446419219*alpha^3
+ 115535699440399080831946228351058377038940*alpha);
n1 := (1/10683)*(-22*d1^4 + 33001*d1^2 - 394);
n2 := (1/10683)*(-22*d2^4 + 33001*d2^2 - 394);
n3 := (1/10683)*(-22*d3^4 + 33001*d3^2 - 394);
// Verify that the coefficients are algebraic integers.
// We verify that the following monic polynomials
// evaluate to zero:
n1^3 - 54*n1^2 + 390*n1 + 13;
n2^3 - 54*n2^2 + 390*n2 + 13;
n3^3 - 54*n3^2 + 390*n3 + 13;
d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
d2^6 - 1488*d2^4 + 4176*d2^2 - 5;
d3^6 - 1488*d3^4 + 4176*d3^2 - 5;
// Now we verify that we have found the exact factorization
// of the desired polynomial:
R<x>:=PolynomialRing(K);
(d1*x - n1)*(d1*x + n1)*(d2*x - n2)*(d2*x + n2)*(d3*x - n3)*(d3*x + n3);
// And the result is 5*x^6 - 144*x^4 + 312*x^2 - 169 !
PARI CODE:
f = y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6 \
+ 28575240325442631201*y^4 - 413298187541059968292500*y^2 \
+ 2460349402142038081821721600
alpha = Mod(y,f)
\\ Define our coefficients:
d1 = (1/1401502976449936390824092894020468134024616320)* \
(- 116858525730451586855252258697*alpha^11 \
+ 21546461434737374316017502217792*alpha^10 \
+ 22109717335034240282753967154849470*alpha^9 \
- 4076616001448173117175959963381396768*alpha^8 \
- 1647209069948549841509121657588818730783*alpha^7 \
+ 303715513906432298157776614831268988666880*alpha^6 \
+ 60404344550512455043182143206734511673314206*alpha^5 \
- 11137524588944779726233177935687516957256627968*alpha^4 \
- 1089475637849489503980551259022753852684824408489*alpha^3 \
+ 200881897616666532603687714008711443626111508690592*alpha^2 \
+ 7719396850213757345324975434862758616298359729245140*alpha\
- 1423343550253117792241514076181168733734615795436123200);
d2 = (1/1401502976449936390824092894020468134024616320)* \
(- 116858525730451586855252258697*alpha^11 \
- 21546461434737374316017502217792*alpha^10 \
+ 22109717335034240282753967154849470*alpha^9 \
+ 4076616001448173117175959963381396768*alpha^8 \
- 1647209069948549841509121657588818730783*alpha^7 \
- 303715513906432298157776614831268988666880*alpha^6 \
+ 60404344550512455043182143206734511673314206*alpha^5 \
+ 11137524588944779726233177935687516957256627968*alpha^4 \
- 1089475637849489503980551259022753852684824408489*alpha^3 \
- 200881897616666532603687714008711443626111508690592*alpha^2 \
+ 7719396850213757345324975434862758616298359729245140*alpha \
+ 1423343550253117792241514076181168733734615795436123200);
d3 = (1/65595009662544996294303701863730606291520)* \
(- 240064724443796467*alpha^11 \
+ 55326448296076857699530*alpha^9 \
- 5309131403735081126863921733*alpha^7 \
+ 274489395789430590995126949847466*alpha^5 \
- 8357164018995647094694127739446419219*alpha^3 \
+ 115535699440399080831946228351058377038940*alpha);
n1 = (1/10683)*(-22*d1^4 + 33001*d1^2 - 394);
n2 = (1/10683)*(-22*d2^4 + 33001*d2^2 - 394);
n3 = (1/10683)*(-22*d3^4 + 33001*d3^2 - 394);
\\ Verify that the coefficients are algebraic integers.
\\ We verify that the following monic polynomials
\\ evaluate to zero:
n1^3 - 54*n1^2 + 390*n1 + 13
n2^3 - 54*n2^2 + 390*n2 + 13
n3^3 - 54*n3^2 + 390*n3 + 13
d1^6 - 1488*d1^4 + 4176*d1^2 - 5
d2^6 - 1488*d2^4 + 4176*d2^2 - 5
d3^6 - 1488*d3^4 + 4176*d3^2 - 5
\\ Now we verify that we have found the exact factorization
\\ of the desired polynomial:
lift( (d1*x-n1)*(d1*x+n1)*(d2*x-n2)*(d2*x+n2)*(d3*x-n3)*(d3*x+n3) )
\\ And the answer is 5*x^6 - 144*x^4 + 312*x^2 - 169 !
Fred the Wonder Worm
In article <3c65f87.03042...@posting.google.com>,
James Harris <jst...@msn.com> wrote:
>
> If you or anyone else can find a factorization of
>
> 5g^6 - 144 g^4 + 312 g^2 - 169
>
> such that
>
> 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
>
> where the c's and d's are algebraic integers, then not only am I wrong
> in disputing your claim, but I don't have proof of Fermat's Last
> Theorem.
Let a be a root of this polynomial:
x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
Define tr, r1, r2, r3 to be the following values:
tr = 176866667462880
r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -
4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -
793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +
4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr
r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -
6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -
459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +
7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr
Note that r1, r2, r3 are all algebraic integers, as they are roots of
the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.
Define ts, s1, s2, s3 to be the following values:
ts = 2947777791048
s1 = (424154940*a^10 + 2120774700*a^9 - 76694478822*a^8 -
319502563488*a^7 + 3300090518376*a^6 + 11027437781076*a^5 -
43851525390876*a^4 - 106459108290348*a^3 + 172083289487202*a^2 +
227204636588880*a - 56537461391712) / ts
s2 = (44176418*a^11 + 30892829*a^10 - 8626342665*a^9 +
2478163251*a^8 + 431167229090*a^7 - 530999629304*a^6 -
7218846139518*a^5 + 14780421135743*a^4 + 38891524066369*a^3 -
99819548314283*a^2 - 30764267126730*a + 151308037861516) / ts
s3 = (-44176418*a^11 - 455047769*a^10 + 6505567965*a^9 +
74216315571*a^8 - 111664665602*a^7 - 2769090889072*a^6 -
3808591641558*a^5 + 29071104255133*a^4 + 67567584223979*a^3 -
72263741172919*a^2 - 196440369462150*a + 64409424246788) / ts
Note that s1, s2, s3 are all algebraic integers, as the are roots of
the (irreducible) monic polynomials x^3 - 54*x^2 + 390*x + 13.
Then set:
c1 = r1, d1 = s1
c2 = r1, d2 = -s1
c3 = r2, d3 = s2
c4 = r2, d4 = -s2
c5 = r3, d5 = s3
c6 = r3, d6 = -s3
and it is the case that:
5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
Cheers,
Geoff.
-----------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@maths.usyd.edu.au | Gameplayer by vocation.
-----------------------------------------------------------------------------
Arturo Magidin
[.snip.]
>Well I'm going to go through and look carefully, but I doubt you went
>to all this trouble to post crap, so I assume you're right, and I'll
>make sure.
>
>Then I'll have no choice but to withdraw my claim of having proven FLT
>as I can't see how I could be right if that polynomial factors with
>algebraic integers.
James, you should realize that you are still wrong: this, by itself,
does not show my theorem is correct, nor does it show your argument is
wrong. The theorem is correct because it has a correct proof. Your
argument is wrong because your divisibility argument is wrong, just as
we have said it is for over a year. There is no connection between the
theorem your challenging and your divisibility result; the
"connection" you have claimed for the past week does not exist.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
I tried e-mail, but of course, it bounced...
>OK, I'll give such a factorization.
Thanks for taking the trouble to do that.
Especially given that there never was a logical connection between the
existence of the factorization and why the divisibility argument James
gave was wrong.
Arturo Magidin
>> In article <210420031722359058%d...@at.dot.com>, Dot <d...@at.dot.com> wrote:
>>
>> I tried e-mail, but of course, it bounced...
>>
>> >OK, I'll give such a factorization.
>>
>> Thanks for taking the trouble to do that.
>>
>> Especially given that there never was a logical connection between the
>> existence of the factorization and why the divisibility argument James
>> gave was wrong.
>
>post that I find fascinating.
>
>How smart are you Magidin?
>
>Oh well, silly question, I'm replying to that guy's post to explain a
>remarkable demonstration that shows that I AM right.
Hmmm... Less than, what, 5 hours, before you start insulting again?
Randy Poe
> ft...@maths.usyd.edu.au (Fred the Wonder Worm) wrote in message news:<b82u36$5aq$1...@spacebar.ucc.usyd.edu.au>...
> > In article <3c65f87.03042...@posting.google.com>,
> > James Harris <jst...@msn.com> wrote:
> > >
> > > If you or anyone else can find a factorization of
> > >
> > > 5g^6 - 144 g^4 + 312 g^2 - 169
> > >
> > > such that
> > >
> > > 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
> > >
> > > where the c's and d's are algebraic integers, then not only am I wrong
> > > in disputing your claim, but I don't have proof of Fermat's Last
> > > Theorem.
>
>
>
> > Let a be a root of this polynomial:
> >
> > x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
> > 488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
> >
I don't have the access to the heavy-duty algebra machines
the others do. I have Maple via Matlab, but I'm not proficient.
So I'll use Matlab's symbolic toolbox to see if I can verify
this with exact arithmetic.
> > Define tr, r1, r2, r3 to be the following values:
> >
> > tr = 176866667462880
> >
> > r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
> > 8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
> > 915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
> > 7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
> >
> > r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -
> > 4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -
> > 793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +
> > 4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr
> >
> > r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -
> > 6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -
> > 459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +
> > 7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr
Matlab:
(I defined r1, r2, r3 symbolically. a has not yet been defined).
>> r1=subs(r1,'tr',tr)
r1 =
986245321/17686666746288*a^11+10848698531/35373333492576*a^10-
195793834069/19651851940320*a^9-2778825491569/58955555820960*a^8+
36817435953311/88433333731440*a^7+18562832062733/11054166716430*a^6-
152641425271111/29477777910480*a^5-3051857657480243/176866667462880*a^4+
3027640866778493/176866667462880*a^3+7747460630371967/176866667462880*a^2+
22435117229609/9825925970160*a-59920475944385/8843333373144
>> r2=subs(r2,'tr',tr)
r2 =
1314760217/35373333492576*a^11+7375768087/44216666865720*a^10-
16783379713/2456481492540*a^9-1455413347249/58955555820960*a^8+
27111005305481/88433333731440*a^7+18593826470359/22108333432860*a^6-
264423249249107/58955555820960*a^5-46474646913544/5527083358215*a^4+
996480374497487/44216666865720*a^3+967878005379331/35373333492576*a^2-
308367050850191/9825925970160*a-283570715694887/8843333373144
>> r3=subs(r3,'tr',tr)
r3 =
1314760217/35373333492576*a^11+42808739587/176866667462880*a^10-
4699074087/727846368160*a^9-140673981019/3684722238810*a^8+
11069599101283/44216666865720*a^7+31049219260843/22108333432860*a^6-
153086084014427/58955555820960*a^5-2663297151231683/176866667462880*a^4+
1065284831907473/176866667462880*a^3+221687448910762/5527083358215*a^2-
4116514893199/409413582090*a-3253396763677/4421666686572
> > Note that r1, r2, r3 are all algebraic integers, as they are roots of
> > the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.
subs('
> Well I'm going to go through and look carefully, but I doubt you went
> to all this trouble to post crap, so I assume you're right, and I'll
> make sure.
>
> Then I'll have no choice but to withdraw my claim of having proven FLT
> as I can't see how I could be right if that polynomial factors with
> algebraic integers.
>
> At that time I'll also do the other that I said I'd do like apologize
> to Magidin for questioning his honesty and mathematical ability, and
> yes, admit to being a fool.
>
> If you know definitely that I'm wrong by being able to check yourself,
> then you already know, but I'm going to check carefully because I
> don't see where the mistake in my reasoning could be. But, hey, I've
> made lots of mistakes before!
>
> Thanks for the effort and the post.
>
>
> James Harris
Randy Poe
> ft...@maths.usyd.edu.au (Fred the Wonder Worm) wrote in message news:<b82u36$5aq$1...@spacebar.ucc.usyd.edu.au>...
> > James Harris <jst...@msn.com> wrote:
> > >
> > > If you or anyone else can find a factorization of
> > >
> > > 5g^6 - 144 g^4 + 312 g^2 - 169
> > >
> > > such that
> > >
> > > 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
> > >
> > > where the c's and d's are algebraic integers, then not only am I wrong
> > > in disputing your claim, but I don't have proof of Fermat's Last
> > > Theorem.
>
Accidentally posted my previous message in which I had begun
a symbolic verification. I didn't know enough about the
symbolic toolbox to finish the job. (What I was hoping to do
was get to a polynomial expression in a, then factor out
the polynomial of which a is known to be z root).
Anyway, here's one other point I wanted to make:
>
> Then I'll have no choice but to withdraw my claim of having proven FLT
> as I can't see how I could be right if that polynomial factors with
> algebraic integers.
>
> At that time I'll also do the other that I said I'd do like apologize
> to Magidin for questioning his honesty and mathematical ability, and
> yes, admit to being a fool.
>
> If you know definitely that I'm wrong by being able to check yourself,
> then you already know, but I'm going to check carefully because I
> don't see where the mistake in my reasoning could be.
There are two glaring errors in reasoning you've been
making. The first is thinking that this theorem had something
to do with your FLT proof. You still believe that. While
your proof has many problems in it, it is not the case that
M-M bears on any particular line of it. Why do you think so?
The second problem is that you think your "arguments" disproving
it have anything to do with the theorem. The statement you're
trying to disprove has to do with coefficients of linear
factors of a polynomial in one variable. You keep going off
with factorizations of other things and not saying anything
at ALL about polynomials in one variable, about linear
factorizations, or about whether or not the coefficients are
algebraic integers.
If you have never, not once, in any of your "arguments" said
that the coefficients can't be algebraic integers, how can
you possibly claim to be disproving something that says they
can be?
- Randy
Virgil
<3c65f87.03042...@posting.google.com>,
jst...@msn.com (James Harris) wrote:
> Let the math speak. Then you can whine AFTER, but you keep whining
> before, so I post here FIRST addressing what you put up FIRST,
> repeatedly.
This must have been written while looking in a mirror.
Who whines most and posts math least?
jst...@msn.com (James Harris)
Virgil
> What's wrong with at least trying to be an adult first?
Why don't you try it for a change, and report back?
Virgil
> Ha! It turns out that "Fred the Wonder Worm" has information in his
> post that I find fascinating.
>
> How smart are you Magidin?
>
> Oh well, silly question, I'm replying to that guy's post to explain a
> remarkable demonstration that shows that I AM right.
>
>
> James Harris
There is none so blind as he who will not see.
Arturo Magidin
>> James Harris <jst...@msn.com> wrote:
>> >
>> > If you or anyone else can find a factorization of
>> >
>> > 5g^6 - 144 g^4 + 312 g^2 - 169
>> >
>> > such that
>> >
>> > 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
>> >
>> > where the c's and d's are algebraic integers, then not only am I wrong
>> > in disputing your claim, but I don't have proof of Fermat's Last
>> > Theorem.
>
>clue in this post and then explain how it proves my case.
>
>> Let a be a root of this polynomial:
>>
>> x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
>> 488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
>>
>> Define tr, r1, r2, r3 to be the following values:
>>
>> tr = 176866667462880
>>
>> r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
>> 8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
>> 915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
>> 7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
>
What makes you think this number is divisible by 25? r1 is an
expression in a, and not all coefficients that show up are multiples
of 25; for example, the leading coefficient 9862453210 is not a
multiple of 25.
Remember that a is a specific number. If all coefficients in that
expression are multiples of some N, then the expression itself is a
multiple of N. However, if not all coefficients are multiples of N,
then sometimes the number will be, and sometimes it will not be.
(Also, you may have forgotten to divide by tr).
>>
>> r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -
>> 4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -
>> 793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +
>> 4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr
>>
>> r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -
>> 6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -
>> 459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +
>> 7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr
>
>And notice here numbers NOT divisible by 25.
Again, this is not immediate; I think you are focusing only on the
last term of the expression, and forgetting the rest.
>> Note that r1, r2, r3 are all algebraic integers, as they are roots of
>> the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.
>
><deleted>
>
>Given that one can now do a neat things as the r's are all factors of
>5 without any other bothersome factors to deal with so I have using r1
Yes. r1*r2*r3 = 5. So each is a divisor of 5 in the ring of all
algebraic integers.
> h*tr* r1 = h*(9862453210*a^11 + 54243492655*a^10 -
>1762144506621*a^9 -
> 8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
> 915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
> 7747460630371967*a^2 + 403832110132962*a - 1198409518887700)
>
>where h is a factor such that h*tr*r1 = ak,
What is k?
r1*tr is the big expression in a:
r1*tr = 9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 +
-8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 +
-915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
7747460630371967*a^2 + 403832110132962*a - 1198409518887700
> so it has a factor that is
>'a',
But this is not a multiple of a, as far as I can see. Unless a happens
to divide 1198409518887700; does it?
So if what you get is a multiple of a, maybe it is because h itself
was what introduced the a... maybe h=a and tr*r1=k?
This seems rather difficult to parse.
> 'a', which remember is a root of
> x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 -
>117533*x^6 - 488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 -
>2931010*x + 987100.
>
>Then multiplying through and collecting gives me
>
> h*9862453210*a^11 + h*54243492655*a^10 - h*1762144506621*a^9 -
> h*8336476474707*a^8 + h*73634871906622*a^7 + h*297005313003728*a^6 -
> h*915848551626666*a^5 - h*3051857657480243*a^4 +
>h*3027640866778493*a^3 + h*7747460630371967*a^2 +
>
> h*403832110132962*a+ ak - h*1198409518887700 = 0
>
>and dividing through by the 5 in the lead coeficient leaves a factor
>of 5 still in the last coefficient.
The last term, -h*1198409518887700 is a multiple of 5; but how do you
know that a(h*403832110132962+k) is a multiple of 5? It is not
obviously so; remember, what you knew was the ri divide 5, not that 5
divides a mulitple of r1.
>But not with the other r's as for instance, with r2, I get
>
>h*6573801085*a^11 + h*29503072348*a^10 - h*1208403339336*a^9 -
> h*4366240041747*a^8 + h*54222010610962*a^7 + h*148750611762872*a^6 -
> h*793269747747321*a^5 - h*1487188701233408*a^4 +
>h*3985921497989948*a^3 +
> h*4839390026896655*a^2 -
>
> h*5550606915303438*a + ak - 5671414313897740 = 0
>
>where dividing off 5 leaves no factors of 5 in the constant term.
I am unclear what it is you think you are accomplishing, or what it is
you were trying to do.
W. Dale Hall
James Harris wrote:
> ft...@maths.usyd.edu.au (Fred the Wonder Worm) wrote in message news:<b82u36$5aq$1...@spacebar.ucc.usyd.edu.au>...
>
>>In article <3c65f87.03042...@posting.google.com>,
>>James Harris <jst...@msn.com> wrote:
>>
>>>If you or anyone else can find a factorization of
>>>
>>> 5g^6 - 144 g^4 + 312 g^2 - 169
>>>
>>>such that
>>>
>>> 5g^6 - 144 g^4 + 312 g^2 - 169 = (c1 g + d1)...(c6 g + d6)
>>>
>>>where the c's and d's are algebraic integers, then not only am I wrong
>>>in disputing your claim, but I don't have proof of Fermat's Last
>>>Theorem.
>>
>
> Ok, it took me a while to see it, but I'll point out an interesting
> clue in this post and then explain how it proves my case.
>
>
And which case would that be?
That is, what is it that you mean to be saying here?
>
> James Harris
Dale
Lukas Horosiewicz
it's getting a bit silly now. You could consider real life conversations
instead of this form.
--
Regards,
Lukas Horosiewicz
"James Harris" <jst...@msn.com> skrev i meddelandet
news:3c65f87.03042...@posting.google.com...
mag...@math.berkeley.edu (Arturo Magidin) wrote in message
news:<b83l6h$onl$1...@agate.berkeley.edu>...
> In article <210420031722359058%d...@at.dot.com>, Dot <d...@at.dot.com>
wrote:
>
> I tried e-mail, but of course, it bounced...
>
> >OK, I'll give such a factorization.
>
> Thanks for taking the trouble to do that.
>
> Especially given that there never was a logical connection between the
> existence of the factorization and why the divisibility argument James
> gave was wrong.
Ha! It turns out that "Fred the Wonder Worm" has information in his
Arturo Magidin
[.snip.]
>> Let a be a root of this polynomial:
>>
>> x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
>> 488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
>>
>> Define tr, r1, r2, r3 to be the following values:
>>
>> tr = 176866667462880
>>
>> r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
>> 8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
>> 915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
>> 7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
>>
>> r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -
>> 4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -
>> 793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +
>> 4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr
>>
>> r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -
>> 6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -
>> 459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +
>> 7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr
>>
>> Note that r1, r2, r3 are all algebraic integers, as they are roots of
>> the (irreducible) monic polynomial x^6 - 1488*x^4 + 4176*x^2 - 5.
>
><deleted>
>
>Using r1
>
> r1*tr = 9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
> 8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
> 915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
> 7747460630371967*a^2 + 403832110132962*a - 1198409518887700
>
>and collecting terms on the right side with a coefficient with a
>factor of 5, gives
>
> r1*tr = 9862453210*a^11 + 54243492655*a^10 - 1198409518887700
>
> - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
>297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
>3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
You have not "collected" anything. You have reordered.
>and now moving to r2
>
> r2*tr = 6573801085*a^11 + 4839390026896655*a^2 - 5671414313897740
>+
>
> 29503072348*a^10 - 1208403339336*a^9 - 4366240041747*a^8 +
>54222010610962*a^7 + 148750611762872*a^6 - 793269747747321*a^5 -
>1487188701233408*a^4 + 3985921497989948*a^3 + 4839390026896655*a^2 -
>5550606915303438*a
>
>where because tr = 176866667462880 both
>
> - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
>297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
>3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
This is the one from r1. Looks right.
>and
>
> 29503072348*a^10 - 1208403339336*a^9 - 4366240041747*a^8 +
>54222010610962*a^7 + 148750611762872*a^6 - 793269747747321*a^5 -
>1487188701233408*a^4 + 3985921497989948*a^3 + 4839390026896655*a^2 -
>5550606915303438*a
>
>must have a factor of 5.
And this is the one from r2.
>But subtracting 403832110132962*a from 5550606915303438*a gives
>
> 5146774805170476a
>
>where 5146774805170476, is, of course, not divisible by 5.
But where did these two numbers come from? They are the last terms of
these expressions; you know that
- 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
is a multiple of 5;
and you know that
29503072348*a^10 - 1208403339336*a^9 - 4366240041747*a^8 +
54222010610962*a^7 + 148750611762872*a^6 - 793269747747321*a^5 -
1487188701233408*a^4 + 3985921497989948*a^3 + 4839390026896655*a^2 -
5550606915303438*a
is a multiple of 5. Why does it follows that subtracting the last
summands will give you a multiple of 5? Certainly, if you take the
entire first thing and subtract the entire second thing will give you
a multiple of 5, but why just the terms
403832110132962*a
and
-5550606915303438*a
?
I mean, in the integers, we have that 3(2)^3 - 2(2)^2 + 2(2) is a multiple
of 5, and we know that 6(2)^3 + 1(2)^2 - 1(2) is a multiple of 5; but that
does not mean that 2(2)-1(2) is a multiple of 5, or that 2(2)+1(2) is
a multiple of 5.
[.rest deleted.]
Arturo Magidin
>
>Using r1
>
> r1*tr = 9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
> 915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
>
>and collecting terms on the right side with a coefficient with a
>factor of 5, gives
>
> r1*tr = 9862453210*a^11 + 54243492655*a^10 - 1198409518887700
>
> - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
>297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
>3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
One usually says "reordering"; unless you meant to write
r1*tr - 9862453210*a^11 - 54243492655*a^10 + 1198409518887700 =
- 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
>and now moving to r2
>
> r2*tr = 6573801085*a^11 + 4839390026896655*a^2 - 5671414313897740
>+
>
> 29503072348*a^10 - 1208403339336*a^9 - 4366240041747*a^8 +
>54222010610962*a^7 + 148750611762872*a^6 - 793269747747321*a^5 -
>1487188701233408*a^4 + 3985921497989948*a^3 + 4839390026896655*a^2 -
>5550606915303438*a
>
>where because tr = 176866667462880 both
>
> - 1762144506621*a^9 - 8336476474707*a^8 + 73634871906622*a^7 +
>297005313003728*a^6 - 915848551626666*a^5 - 3051857657480243*a^4 +
>3027640866778493*a^3 + 7747460630371967*a^2 + 403832110132962*a
>
>and
>
> 29503072348*a^10 - 1208403339336*a^9 - 4366240041747*a^8 +
>54222010610962*a^7 + 148750611762872*a^6 - 793269747747321*a^5 -
>1487188701233408*a^4 + 3985921497989948*a^3 + 4839390026896655*a^2 -
>5550606915303438*a
>
>must have a factor of 5.
>
>But subtracting 403832110132962*a from 5550606915303438*a gives
>
> 5146774805170476a
>
>where 5146774805170476, is, of course, not divisible by 5.
>
>Then, in any case where 'a' itself has a factor of 5, that factor MUST
>be 5.
Why have you isolated the last terms of these expressions? What makes
you think that (5550606915303438*a - 403832110132962*a) will
necessarily be a multiple of 5?
In the integers, with a value of 2, we have:
3(2)^3 - 2(2)^2 + 2(2) = 20 a multiple of 5 (none of the coefficients,
3, -2, 2 are multiples of 5)
and
6(2)^3 + 1(2)^2 - (2) = 50 a multiple of 5 (none of the coefficients,
6, 1, -2 multiples of 5),
but neither 2(2)-(2)=2 nor 2(2)+2=6 are multiples of 5. You can
certainly say that
(3-6)(2^3) + (-2-1)(2)^2 + (2-(-1))(2) = 20-50=-30 is a multiple of 5
(by taking the difference of the full expressions), but not just the
last terms.
Unless you have some other justification for why 5146774805170476a is
supposed to be divisible by 5?
Steven Lord
*snip*
James, James, James. If I remember correctly, you'd originally wanted your
proof of FLT to be a simple one, one that could be taught in middle or high
schools. As it stands now, I'm not sure it satisfies that goal -- it's too
complicated for that purpose. I think that you should take a break for a
day or two and think about how you can eliminate some of the complexity of
the proof. Then, start writing it up, explaining everything as simply as
you can. Once you've finished, before exposing it to the people on this
group who obviously don't recognize what you've done, find a local teenager
and let them read it. If they understand it, great. If they can't, rewrite
the sections that they find confusing. Then, once you have the simple proof
in a simple form, post it and proclaim that a teenager could understand it.
Wouldn't that embarass the people who claim that it's wrong if you did that
and they still didn't understand it?
Steve L
Victor Eijkhout
> I think that you should take a break for a
> day or two
s/day/year/
V.
--
mail me at lastname at cs utk edu
Randy Poe
> Were you trying to keep me off-balance so that I wouldn't be able to
> see that nifty result about factors of 5?
>
That "nifty result" is as fictitious as any of your other
observations about "factors". You have been looking at an
expression which is of the form "r = (polynomial in a)/tr"
where a and tr are algebraic integers. You are claiming that
you can factor 5 out of the polynomial, therefore 5 is
a "factor" of this number.
In what sense? Again you're reaching, as you have for years,
for a definition of "factor" which is by inspection and not
in terms of multiplication. And therefore it doesn't have the
properties of a factor.
When you pull a 5 out of that thing, the thing that is left
is NOT guaranteed to be an algebraic integer. r is an algebraic
integer, but r/5 may not be. Probably isn't.
I believe that further you have extracted sub-polynomials,
and tried to claim they are meaningful. You're writing
r = (p1 + p2)/tr and trying to say that p2/tr has a "factor
of 5". Not only don't you know if p2/(5*tr) is an algebraic
integer, you don't know if p2/tr is.
This will not be new to you, but you seem to have never absorbed
its meaning: We say that a is a factor of c (in some ring) if
we can write c = a*b, where a, c, AND b are all members of
that ring.
What ring is r/5 a member of? What ring is p2/tr a member of?
What ring is p2/(5*tr) a member of? Don't use the word "factor"
unless you're willing to say what ring you're factoring over.
Otherwise it's meaningless. You've heard that many times. But
still you ignore it.
- Randy
Virgil
> Well Dot all that excess in your post wasn't necessary.
Do you now have the gall to pass YOURSELF off as a model of
posting only what is necessary?
The ratio of emotional excess to useful communication in
your average posting is immense.
Whydon't you try to get that ratio down to double digits?
Dot
> Well Dot all that excess in your post wasn't necessary. [...]
Those extra 10 lines you mean? Two of which mentioned where to
get math software that would allow you to verify my claims? Sorry.
> Were you trying to keep me off-balance so that I wouldn't be able to
> see that nifty result about factors of 5? [...]
> Is that why you gave a *numerical* result such emphasis?
I started out with the numerical results for the reasons I stated:
I though they would be easier for you to verify without having to
resort to math software that I didn't think you had. But in this
post I will give exact results, like you say you want, because I am
hoping that you will answer a simple question (at the end of the post)
about your divisibility claims.
Let me see if I can summarize some of your contentions from various
threads, having to do with your "exactly two of the a's have factors
of sqrt(5)" arguments.
Write x^3 - 12*x^2 + 65 = (x - a1) * (x - a2) * (x - a3).
Then you say that exactly two of the a's are divisible by sqrt(5),
correct? It would then follow that exactly two of the a^2's would
be divisible by 5, right? I think that some of the algebra would
be simpler without that square root, so let's figure out a polynomial
satisfied by the a^2's (as I believe you yourself, and others,
have done before).
Since the a's satisfy a^3 - 12*a^2 + 65 = 0, we have
a^3 = (12*a^2 - 65)
a^6 = 144*a^4 - 1560*a^2 + 4225
(a^2)^3 - 144*(a^2)^2 + 1560*(a^2) - 4225 = 0
So we have
x^3 - 144*x^2 + 1560*x - 4225 = (x - a1^2) * (x - a2^2) * (x - a3^2).
Write c1 = a1^2, etc. so that we have
x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3).
Then you say that exactly two of the c's are divisible by 5,
right? OK, so let's write out what the c's are. Let r be a root of
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
and then take
c1 := (1/27857196)*(- 132109*r^5 - 25904074*r^4 + 1069057055*r^3
+ 1250725412*r^2 - 3528409195*r + 208158188);
c2 := (1/27857196)*( 30227*r^5 + 5884541*r^4 - 252808609*r^3
+ 78301427*r^2 + 288642953*r + 186689843);
c3 := (1/27857196)*( 101882*r^5 + 20019533*r^4 - 816248446*r^3
- 1329026839*r^2 + 3239766242*r + 3616588193);
You can check that then
x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3)
so these really are the right c's.
Here's my question: You claim that two of the c's are divisible
by 5 and that one is not. So tell me please, which is which?
-- Dot.
James Harris
<deleted>
I'm guessing that it's c2 and c3 and here's how you prove that c1 does
NOT have any non unit algebraic integer factors of 5, if I'm right.
Take
P(r) = 132109*r^5 - 25904074*r^4 + 1250725412*r^2 + 208158188
from your results for c1, and note that it must have any non unit
algebraic integer factors of 5 shared with c1, if there are any.
Now set your polynomial
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
and make the substitution with r, then multiply by 132109 and divide
by P(r) and take the remainder.
Now divide P(r) by that result and take the remainder, which *should*
be an integer which does NOT have a factor of 5.
Algebra IS fundamental.
James Harris
Dot
In a post other than the one to which I am responding you complain
about some extra-mathematical comments I made recently. Well, you
are perfectly right: I was not polite, and I should have been.
> > Then you say that exactly two of the c's are divisible by 5,
> > right? OK, so let's write out what the c's are. Let r be a root of
> >
> > x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> >
> > and then take
> >
> > c1 := (1/27857196)*(- 132109*r^5 - 25904074*r^4 + 1069057055*r^3
> > + 1250725412*r^2 - 3528409195*r + 208158188);
> > c2 := (1/27857196)*( 30227*r^5 + 5884541*r^4 - 252808609*r^3
> > + 78301427*r^2 + 288642953*r + 186689843);
> > c3 := (1/27857196)*( 101882*r^5 + 20019533*r^4 - 816248446*r^3
> > - 1329026839*r^2 + 3239766242*r + 3616588193);
> >
> > You can check that then
> >
> > x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3)
> >
> > so these really are the right c's.
> >
> > Here's my question: You claim that two of the c's are divisible
> > by 5 and that one is not. So tell me please, which is which?
> >
> > -- Dot.
>
> I'm guessing that it's c2 and c3 and here's how you prove that c1 does
> NOT have any non unit algebraic integer factors of 5, if I'm right.
>
> Take
>
> P(r) = 132109*r^5 - 25904074*r^4 + 1250725412*r^2 + 208158188
(There should be a minus sign in front of the 132109*r^5 term.)
>
> from your results for c1, and note that it must have any non unit
> algebraic integer factors of 5 shared with c1, if there are any.
>
> Now set your polynomial
>
> x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
>
> and make the substitution with r, then multiply by 132109 and divide
> by P(r) and take the remainder.
>
> Now divide P(r) by that result and take the remainder, which *should*
> be an integer which does NOT have a factor of 5.
Careful! You have to make a distinction between polynomials and values
that polynomials take when you substitute values in. But with the right
interpretation, the argument you outline could actually work. You have
to go through a few more iterations of the process to get an integer
and not a polynomial, but the idea is sound. The number you get at the
end is called the "resultant" of the two polynomials you started with.
But as it happens the resultant of
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
and
-132109*x^5 - 25904074*x^4 + 1250725412*x^2 + 208158188
is
-2691497186477896891141888811255638554956355038118698285853342899375,
which is clearly divisible by 5. So you can't show that c1 is not
divisible by 5 this way. (On the other hand, this also doesn't show
that c1 *is* divisible by 5.)
In fact, the computation works out a little bit nicer if you *don't*
throw away parts of the expression for c1. That is, suppose you take
the resultant of
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
and
- 132109*x^5 - 25904074*x^4 + 1069057055*x^3
+ 1250725412*x^2 - 3528409195*x + 208158188.
The answer is
8342146754510965740185274171990390295421810588160000,
again a multiple of 5.
In fact, if you go through the same procedure with c2 and c3, your
get 8342146754510965740185274171990390295421810588160000 each time.
So the argument you suggest cannot show that *any* of the c's is not
a multiple of 5.
-- Dot.
Dot
In article <3c65f87.03042...@posting.google.com>, James
Harris <jst...@msn.com> wrote:
> Dot <d...@at.dot.com> wrote in message
> news:<240420030049317148%d...@at.dot.com>...
[snip]
> > Then you say that exactly two of the c's are divisible by 5,
> > right? OK, so let's write out what the c's are. Let r be a root of
> >
> > x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> >
> > and then take
> >
> > c1 := (1/27857196)*(- 132109*r^5 - 25904074*r^4 + 1069057055*r^3
> > + 1250725412*r^2 - 3528409195*r + 208158188);
> > c2 := (1/27857196)*( 30227*r^5 + 5884541*r^4 - 252808609*r^3
> > + 78301427*r^2 + 288642953*r + 186689843);
> > c3 := (1/27857196)*( 101882*r^5 + 20019533*r^4 - 816248446*r^3
> > - 1329026839*r^2 + 3239766242*r + 3616588193);
> >
> > You can check that then
> >
> > x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3)
> >
> > so these really are the right c's.
> >
> > Here's my question: You claim that two of the c's are divisible
> > by 5 and that one is not. So tell me please, which is which?
> >
> > -- Dot.
>
> I'm guessing that it's c2 and c3 and here's how you prove that c1 does
> NOT have any non unit algebraic integer factors of 5, if I'm right.
>
> [argument snipped, and addressed in a different followup.]
James,
For the sake of the argument, suppose that you are correct and that c1
is not divisible by 5 and yet c2 and c3 are divisible by 5. Now,
whatever reasoning leads you to this conclusion surely does not depend
on the fact that we called the variables r and c1, c2, c3, right? So
if you have an argument showing that c1 is not divisible by 5, then the
*exact* *same* *argument* would show the following:
Let s be a root of
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
(the same polynomial that r is a root of) and then take
d1 := (1/27857196)*(- 132109*s^5 - 25904074*s^4 + 1069057055*s^3
+ 1250725412*r^2 - 3528409195*r + 208158188);
d2 := (1/27857196)*( 30227*s^5 + 5884541*s^4 - 252808609*s^3
+ 78301427*s^2 + 288642953*s + 186689843);
d3 := (1/27857196)*( 101882*s^5 + 20019533*s^4 - 816248446*s^3
- 1329026839*s^2 + 3239766242*s + 3616588193);
(so that the relationship of the d's to s is the same as the
relationship of the c's to r). Then your argument would show that
d1 is not a multiple of 5 and that d2 and d3 are multiples of 5.
Do you agree? I am assuming that you will.
So now I will choose an s and apply your argument. I will choose
s := (1/1086430644)*( 9641057*r^5 + 1889894450*r^4
- 78120717223*r^3 - 86585494222*r^2
+ 253854465581*r - 4315626502);
where r is the same r as before. Now, the first funny thing is that
this s really is also a root of
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569,
as is required. And the second funny thing is that with this choice
of r and s, and with the definitions of the c's and d's given above,
we have
c1 = d2 and c2 = d1 and c3 = d3.
(If you have the appropriate software, you really should check this!)
So your argument that c1 is not divisible by 5, if it is valid, would
show that d1 is not divisible by 5, and since c2 = d1 it would follow
that c2 is not divisible by 5. Likewise, if you could show that c2 is
divisible by 5, then the same argument would show that d2 is divisible
by 5, which would show that c1 is divisible by 5.
So either *both* c1 and c2 are divisible by 5, or *neither* of them is.
How can this gibe with your claim that exactly two of the c's are
divisible by 5? I suppose you could hope that maybe c1 and c2 are
divisible by 5 and that c3 is not, but trust me, in that case I will
provide you with a t and an e1, e2, e3, such that...
-- Dot.
James Harris
> James,
>
> In a post other than the one to which I am responding you complain
> about some extra-mathematical comments I made recently. Well, you
> are perfectly right: I was not polite, and I should have been.
Ok. Thanks.
> > > Then you say that exactly two of the c's are divisible by 5,
> > > right? OK, so let's write out what the c's are. Let r be a root of
> > >
> > > x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> > >
> > > and then take
> > >
> > > c1 := (1/27857196)*(- 132109*r^5 - 25904074*r^4 + 1069057055*r^3
> > > + 1250725412*r^2 - 3528409195*r + 208158188);
> > > c2 := (1/27857196)*( 30227*r^5 + 5884541*r^4 - 252808609*r^3
> > > + 78301427*r^2 + 288642953*r + 186689843);
> > > c3 := (1/27857196)*( 101882*r^5 + 20019533*r^4 - 816248446*r^3
> > > - 1329026839*r^2 + 3239766242*r + 3616588193);
> > >
> > > You can check that then
> > >
> > > x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3)
> > >
> > > so these really are the right c's.
> > >
> > > Here's my question: You claim that two of the c's are divisible
> > > by 5 and that one is not. So tell me please, which is which?
> > >
> > > -- Dot.
> >
<deleted>
> >
> > Now set your polynomial
> >
> > x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> >
> > and make the substitution with r, then multiply by 132109 and divide
> > by P(r) and take the remainder.
> >
> > Now divide P(r) by that result and take the remainder, which *should*
> > be an integer which does NOT have a factor of 5.
>
> Careful! You have to make a distinction between polynomials and values
> that polynomials take when you substitute values in. But with the right
> interpretation, the argument you outline could actually work. You have
> to go through a few more iterations of the process to get an integer
> and not a polynomial, but the idea is sound. The number you get at the
> end is called the "resultant" of the two polynomials you started with.
>
> But as it happens the resultant of
> x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> and
> -132109*x^5 - 25904074*x^4 + 1250725412*x^2 + 208158188
> is
> -2691497186477896891141888811255638554956355038118698285853342899375,
> which is clearly divisible by 5. So you can't show that c1 is not
> divisible by 5 this way. (On the other hand, this also doesn't show
> that c1 *is* divisible by 5.)
Ok.
> In fact, the computation works out a little bit nicer if you *don't*
> throw away parts of the expression for c1. That is, suppose you take
> the resultant of
> x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> and
> - 132109*x^5 - 25904074*x^4 + 1069057055*x^3
> + 1250725412*x^2 - 3528409195*x + 208158188.
> The answer is
> 8342146754510965740185274171990390295421810588160000,
> again a multiple of 5.
>
> In fact, if you go through the same procedure with c2 and c3, your
> get 8342146754510965740185274171990390295421810588160000 each time.
> So the argument you suggest cannot show that *any* of the c's is not
> a multiple of 5.
>
> -- Dot.
Hmmm...it seems to me there should be a way with your example but I'll
have to think more on it. Still, note that your example here is not
necessary for my case as I can use a previous post by you where the
key difference is at least some of the r's must have non unit
algebraic integer factors of 5.
Now if I made some error in my reasoning with your previous results
that's more interesting to me.
But I'm also going to keep thinking about this example.
James Harris
James Harris
No, it does not so depend.
>So
> if you have an argument showing that c1 is not divisible by 5, then the
> *exact* *same* *argument* would show the following:
>
> Let s be a root of
>
> x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
>
> (the same polynomial that r is a root of) and then take
>
> d1 := (1/27857196)*(- 132109*s^5 - 25904074*s^4 + 1069057055*s^3
> + 1250725412*r^2 - 3528409195*r + 208158188);
> d2 := (1/27857196)*( 30227*s^5 + 5884541*s^4 - 252808609*s^3
> + 78301427*s^2 + 288642953*s + 186689843);
> d3 := (1/27857196)*( 101882*s^5 + 20019533*s^4 - 816248446*s^3
> - 1329026839*s^2 + 3239766242*s + 3616588193);
>
> (so that the relationship of the d's to s is the same as the
> relationship of the c's to r). Then your argument would show that
> d1 is not a multiple of 5 and that d2 and d3 are multiples of 5.
>
>
> Do you agree? I am assuming that you will.
Why would I agree?
If I have x^2 + 2x + 1 as a polynomial and then take a root r, and a
root s, and then write
r^2 + 2s + 1 = 0
would you agree?
It can be true, but only if r=s.
> So now I will choose an s and apply your argument. I will choose
>
> s := (1/1086430644)*( 9641057*r^5 + 1889894450*r^4
> - 78120717223*r^3 - 86585494222*r^2
> + 253854465581*r - 4315626502);
>
> where r is the same r as before.
You can't say the "same r as before" as r has multiple solutions.
Think of it as just like you can have
P(x) = 0, with multiple roots,
you can have
P(x) = 5, with multiple values for x,
or you can have
P(x) = g, where g is an algebraic integer
which is the situation.
The only difference is that instead of zeroing the polynomial, you're
setting it to some value, and for the d's that value is an algebraic
integer.
Which is why there are three different polynomials for the d's.
In each case though, as you use the roots r, you get the same answer
from each polynomial.
So you can use each root from
x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
and get the same value for each of the d's no matter which roots you
use.
>Now, the first funny thing is that
> this s really is also a root of
>
> x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569,
>
> as is required. And the second funny thing is that with this choice
> of r and s, and with the definitions of the c's and d's given above,
> we have
>
> c1 = d2 and c2 = d1 and c3 = d3.
>
> (If you have the appropriate software, you really should check this!)
> So your argument that c1 is not divisible by 5, if it is valid, would
> show that d1 is not divisible by 5, and since c2 = d1 it would follow
> that c2 is not divisible by 5. Likewise, if you could show that c2 is
> divisible by 5, then the same argument would show that d2 is divisible
> by 5, which would show that c1 is divisible by 5.
>
> So either *both* c1 and c2 are divisible by 5, or *neither* of them is.
>
> How can this gibe with your claim that exactly two of the c's are
> divisible by 5? I suppose you could hope that maybe c1 and c2 are
> divisible by 5 and that c3 is not, but trust me, in that case I will
> provide you with a t and an e1, e2, e3, such that...
>
> -- Dot.
Why don't we just use your other results?
There I can show you *mathematically* with simple algebra exactly the
situation you're wondering about, and then the interesting discussion
about how it's possible can take place.
But I see that you're still doubtful that it can happen, so I think
that a first step would be using your other results, and then coming
back to this example to puzzle over it.
James Harris
James Harris
> In article <3c65f87.03042...@posting.google.com>, James
> Harris <jst...@msn.com> wrote:
>
> > Dot <d...@at.dot.com> wrote in message
> > news:<240420030049317148%d...@at.dot.com>...
>
> [snip]
>
> > > Then you say that exactly two of the c's are divisible by 5,
> > > right? OK, so let's write out what the c's are. Let r be a root of
> > >
> > > x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
> > >
> > > and then take
> > >
> > > c1 := (1/27857196)*(- 132109*r^5 - 25904074*r^4 + 1069057055*r^3
> > > + 1250725412*r^2 - 3528409195*r + 208158188);
> > > c2 := (1/27857196)*( 30227*r^5 + 5884541*r^4 - 252808609*r^3
> > > + 78301427*r^2 + 288642953*r + 186689843);
> > > c3 := (1/27857196)*( 101882*r^5 + 20019533*r^4 - 816248446*r^3
> > > - 1329026839*r^2 + 3239766242*r + 3616588193);
> > >
> > > You can check that then
> > >
> > > x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3)
> > >
> > > so these really are the right c's.
> > >
> > > Here's my question: You claim that two of the c's are divisible
> > > by 5 and that one is not. So tell me please, which is which?
> > >
> > > -- Dot.
<deleted>
I thought I'd copy from your other post and explain in detail why your
own data answers the question you raise here.
Here's a quote from your post
messageid: <210420032353503610%d...@at.dot.com>
<Quote>
We will work in the number field defined by the polynomial
y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
+ 28575240325442631201*y^4 - 413298187541059968292500*y^2
+ 2460349402142038081821721600;
We will let alpha be a root of this polynomial, and we will define
algebraic integers d1, d2, d3, n1, n2, n3 in terms of alpha.
(Taking alpha to be the real root that is approximately
154.451496990997802982236692688237765859998280922
gives the numerical examples I gave earlier.
MAGMA CODE:
R<y>:=PolynomialRing(Rationals());
f := y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
+ 28575240325442631201*y^4 - 413298187541059968292500*y^2
+ 2460349402142038081821721600;
K<alpha>:=NumberField(f);
// Define our coefficients:
d1 := (1/1401502976449936390824092894020468134024616320)*
(- 116858525730451586855252258697*alpha^11
+ 21546461434737374316017502217792*alpha^10
+ 22109717335034240282753967154849470*alpha^9
- 4076616001448173117175959963381396768*alpha^8
- 1647209069948549841509121657588818730783*alpha^7
+ 303715513906432298157776614831268988666880*alpha^6
+ 60404344550512455043182143206734511673314206*alpha^5
- 11137524588944779726233177935687516957256627968*alpha^4
- 1089475637849489503980551259022753852684824408489*alpha^3
+ 200881897616666532603687714008711443626111508690592*alpha^2
+ 7719396850213757345324975434862758616298359729245140*alpha
- 1423343550253117792241514076181168733734615795436123200);
d2 := (1/1401502976449936390824092894020468134024616320)*
(- 116858525730451586855252258697*alpha^11
- 21546461434737374316017502217792*alpha^10
+ 22109717335034240282753967154849470*alpha^9
+ 4076616001448173117175959963381396768*alpha^8
- 1647209069948549841509121657588818730783*alpha^7
- 303715513906432298157776614831268988666880*alpha^6
+ 60404344550512455043182143206734511673314206*alpha^5
+ 11137524588944779726233177935687516957256627968*alpha^4
- 1089475637849489503980551259022753852684824408489*alpha^3
- 200881897616666532603687714008711443626111508690592*alpha^2
+ 7719396850213757345324975434862758616298359729245140*alpha
+ 1423343550253117792241514076181168733734615795436123200);
d3 := (1/65595009662544996294303701863730606291520)*
(- 240064724443796467*alpha^11
+ 55326448296076857699530*alpha^9
- 5309131403735081126863921733*alpha^7
+ 274489395789430590995126949847466*alpha^5
- 8357164018995647094694127739446419219*alpha^3
+ 115535699440399080831946228351058377038940*alpha);
n1 := (1/10683)*(-22*d1^4 + 33001*d1^2 - 394);
n2 := (1/10683)*(-22*d2^4 + 33001*d2^2 - 394);
n3 := (1/10683)*(-22*d3^4 + 33001*d3^2 - 394);
// Verify that the coefficients are algebraic integers.
// We verify that the following monic polynomials
// evaluate to zero:
n1^3 - 54*n1^2 + 390*n1 + 13;
n2^3 - 54*n2^2 + 390*n2 + 13;
n3^3 - 54*n3^2 + 390*n3 + 13;
d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
d2^6 - 1488*d2^4 + 4176*d2^2 - 5;
d3^6 - 1488*d3^4 + 4176*d3^2 - 5;
</Quote>
Ok, I'll use your d1, which is
d1 := (1/1401502976449936390824092894020468134024616320)*
(- 116858525730451586855252258697*alpha^11
+ 21546461434737374316017502217792*alpha^10
+ 22109717335034240282753967154849470*alpha^9
- 4076616001448173117175959963381396768*alpha^8
- 1647209069948549841509121657588818730783*alpha^7
+ 303715513906432298157776614831268988666880*alpha^6
+ 60404344550512455043182143206734511673314206*alpha^5
- 11137524588944779726233177935687516957256627968*alpha^4
- 1089475637849489503980551259022753852684824408489*alpha^3
+ 200881897616666532603687714008711443626111508690592*alpha^2
+ 7719396850213757345324975434862758616298359729245140*alpha
- 1423343550253117792241514076181168733734615795436123200);
so
1401502976449936390824092894020468134024616320 d1 :=
+ 22109717335034240282753967154849470*alpha^9
+ 303715513906432298157776614831268988666880*alpha^6
+ 7719396850213757345324975434862758616298359729245140*alpha
- 1423343550253117792241514076181168733734615795436123200
- 116858525730451586855252258697*alpha^11
+ 21546461434737374316017502217792*alpha^10
- 4076616001448173117175959963381396768*alpha^8
- 1647209069948549841509121657588818730783*alpha^7
+ 60404344550512455043182143206734511673314206*alpha^5
- 11137524588944779726233177935687516957256627968*alpha^4
- 1089475637849489503980551259022753852684824408489*alpha^3
+ 200881897616666532603687714008711443626111508690592*alpha^2
);
Where I've grouped terms with a coefficient with a factor of 5.
From there I get
1401502976449936390824092894020468134024616320 d1 :=
22109717335034240282753967154849470*alpha^9
+ 303715513906432298157776614831268988666880*alpha^6
+ 7719396850213757345324975434862758616298359729245140*alpha
- 1423343550253117792241514076181168733734615795436123200
(alpha^2)(- 116858525730451586855252258697*alpha^9
+ 21546461434737374316017502217792*alpha^8
- 4076616001448173117175959963381396768*alpha^6
- 1647209069948549841509121657588818730783*alpha^5
+ 60404344550512455043182143206734511673314206*alpha^3
- 11137524588944779726233177935687516957256627968*alpha^2
- 1089475637849489503980551259022753852684824408489*alpha^1
+ 200881897616666532603687714008711443626111508690592)
);
and letting
j = 22109717335034240282753967154849470*alpha^9
+ 303715513906432298157776614831268988666880*alpha^6
+ 7719396850213757345324975434862758616298359729245140*alpha
- 1423343550253117792241514076181168733734615795436123200
I have that j/5 still would have non unit algebraic integer factors of
5 because each of the coefficients next to an alpha has a factor of 5,
so the alpha's would still be left, and the last coefficient
- 1423343550253117792241514076181168733734615795436123200
has a factor of 25, so it'd have a factor of 5 left.
And I would divide out but I'd have to write a quick program to do so
with such big numbers and I would rather just let people look at the
last two digits.
Now letting
k = (- 116858525730451586855252258697*alpha^9
+ 21546461434737374316017502217792*alpha^8
- 4076616001448173117175959963381396768*alpha^6
- 1647209069948549841509121657588818730783*alpha^5
+ 60404344550512455043182143206734511673314206*alpha^3
- 11137524588944779726233177935687516957256627968*alpha^2
- 1089475637849489503980551259022753852684824408489*alpha^1
+ 200881897616666532603687714008711443626111508690592)
I note that k does not have factors of 5, if alpha does because of the
coefficient
200881897616666532603687714008711443626111508690592
and now I can write
1401502976449936390824092894020468134024616320 d1 := j + alpha^2 k
and consider that because
1401502976449936390824092894020468134024616320
has a factor of 5, that is 5, of course, I can divide that off and
consider
j/5 + (alpha^2 k)/5
and I've already noted above that j/5 still has any non unit algebraic
integer factors of alpha, so now (alpha^2 k)/5 would be forced to have
any non unit algebraic integer factors of d1, if it has any, and
remember that k does not have any non unit algebraic integer factors
of 5, if alpha does.
And remember because alpha is a root of
f := y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
+ 28575240325442631201*y^4 - 413298187541059968292500*y^2
+ 2460349402142038081821721600;
it must hae solutions that have non unit algebraic integer factors of
5.
However, consider what happens when we use a root alpha to get
(alpha)^12 - 226446*(alpha)^10 + 21142596231*(alpha)^8 -
1041903035971246*(alpha)^6 + 28575240325442631201*(alpha)^4 -
413298187541059968292500*(alpha)^2
+ 2460349402142038081821721600 = 0
and notice that the last coefficient has a factor that is 25 and then
notice the coefficient of (alpha)^2 has a factor of 25 as well.
and if you divide through by alpha^2 you get
(alpha)^10 - 226446*(alpha)^8 + 21142596231*(alpha)^6 -
1041903035971246*(alpha)^4 + 28575240325442631201*(alpha)^2 -
413298187541059968292500
+ 2460349402142038081821721600/(alpha^2) = 0
so if there exists an algebraic integer h, such that
alpha^2 h = 25
where alpha and h are coprime to each other than the last term would
no longer have factors of 5 in alpha.
It turns out that
alpha^2 h = 5
must be true, as anything greater, like
alpha^2 h = 5 sqrt(5)
would still create a contradiction, but from before I have
j/5 + (alpha^2 k)/5
and now I see that it cannot have non unit algebraic integer factors
of 5, which proves that d1 does not either.
Now notice that the same argument works for d2, but d3 is different,
as the same approach proves that it does share factors of 5 with
alpha.
That proves that
d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
has unit algebraic integer roots.
And Dot that answers your question above.
James Harris
Dot
wrote:
> d1 := (1/27857196)*(- 132109*s^5 - 25904074*s^4 + 1069057055*s^3
> + 1250725412*r^2 - 3528409195*r + 208158188);
> d2 := (1/27857196)*( 30227*s^5 + 5884541*s^4 - 252808609*s^3
> + 78301427*s^2 + 288642953*s + 186689843);
> d3 := (1/27857196)*( 101882*s^5 + 20019533*s^4 - 816248446*s^3
> - 1329026839*s^2 + 3239766242*s + 3616588193);
Aack. The expression for d1 should contain no r's. Those should all
be s's.
-- Dot.
James Harris
That's ok. In case you missed it, here's the analysis that answers
questions you've raised. And proves irrefutably a rather nifty result
in number theory.
messageid: <210420032353503610%d...@at.dot.com>
MAGMA CODE:
R<y>:=PolynomialRing(Rationals());
K<alpha>:=NumberField(f);
// Define our coefficients:
</Quote>
so
1401502976449936390824092894020468134024616320 d1 :=
- 116858525730451586855252258697*alpha^11
+ 21546461434737374316017502217792*alpha^10
From there I get
1401502976449936390824092894020468134024616320 d1 :=
(alpha^2)(- 116858525730451586855252258697*alpha^9
+ 21546461434737374316017502217792*alpha^8
and letting
I have that j/5 still would have non unit algebraic integer factors of
5 as long as alpha does because each of the coefficients next to an
- 1423343550253117792241514076181168733734615795436123200
Now letting
200881897616666532603687714008711443626111508690592
and consider that because
1401502976449936390824092894020468134024616320
j/5 + (alpha^2 k)/5
alpha^2 h = 25
It turns out that
alpha^2 h = 5
j/5 + (alpha^2 k)/5
That proves that
James Harris
James Harris
Sorry it took a while Dot, but I can finally answer that question with
a demonstration in addition to the mathematical argument that proves
it which I've given in various forms now for about a year.
But I understand that many of you don't believe in mathematics itself
enough to simply accept a proof when a mathematician like Magidin is
arguing against it, even when he can't give a valid error in the
proof, as after all, proofs don't have errors, so he couldn't give an
error. But he could get people to believe him, if they didn't believe
in logic more than in a human being.
Math is HARD because it depends on a belief in logic. And lots of
people just don't believe in logic, as I've discovered.
Your question is answered rather easily by considering the polynomial
d^3 - 1488*d^2 + 4176*d - 5
directly as amazingly, in a bit of luck I guess,
d(d^2 + 2d + 1 - 1490d + 4175) = d^3 - 1488*d^2 + 4176*d
so
d(d^2 + 2d + 1 - 1490d + 4175) = 5, which means
d[{d + 1)^2 - 1490d + 4175] = 5
which forces any root plus 1 to have the algebraic integer factors of
any other root that is not coprime to 5.
But looking at the solutions to
d^3 - 1488*d^2 + 4176*d - 5 = 0
with that information proves that at least one root must be coprime to
5.
Those solutions are
{{x -> 496 + 244624/(1/2*(241976581 +
10683*I*Sqrt[12415]))^(1/3) +
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
{x -> 496 - (122312*(1 + I*Sqrt[3]))/
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
1/2*(1 - I*Sqrt[3])*
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
{x -> 496 - (122312*(1 - I*Sqrt[3]))/
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
1/2*(1 + I*Sqrt[3])*
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}}
where the last two subtracted from each other give
244624/(1/2*(241976581 +
10683*I*Sqrt[12415]))^(1/3) +
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}
times I*Sqrt[3] proving that
496 + 244624/(1/2*(241976581 +
10683*I*Sqrt[12415]))^(1/3) +
(1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
must be coprime to 5.
Note that I've been forced to give demonstrations such as this for a
PROOF that I've had for about a year now.
I think it unfair and unfortunate that mathematics itself wasn't
accepted, and I fear that even a secondary demonstration like this one
may not sway mathematicians.
I did my best. I found a short proof of Fermat's Last Theorem, and
then people wanted more.
Now I've given more, and I don't know if it will be enough.
James Harris
Arturo Magidin
James Harris <jst...@msn.com> wrote:
[.snip.]
>> Then you say that exactly two of the c's are divisible by 5,
>> right? OK, so let's write out what the c's are. Let r be a root of
>>
>> x^6 + 197*x^5 - 7912*x^4 - 16891*x^3 + 18248*x^2 + 24437*x - 569
>>
>> and then take
>>
>> c1 := (1/27857196)*(- 132109*r^5 - 25904074*r^4 + 1069057055*r^3
>> + 1250725412*r^2 - 3528409195*r + 208158188);
>> c2 := (1/27857196)*( 30227*r^5 + 5884541*r^4 - 252808609*r^3
>> + 78301427*r^2 + 288642953*r + 186689843);
>> c3 := (1/27857196)*( 101882*r^5 + 20019533*r^4 - 816248446*r^3
>> - 1329026839*r^2 + 3239766242*r + 3616588193);
>>
>> You can check that then
>>
>> x^3 - 144*x^2 + 1560*x - 4225 = (x - c1) * (x - c2) * (x - c3)
>>
>> so these really are the right c's.
>>
>> Here's my question: You claim that two of the c's are divisible
>> by 5 and that one is not. So tell me please, which is which?
[.snip.]
>Your question is answered rather easily by considering the polynomial
It's interesting. You may be answering ->a<- question, but you are NOT
answering Dot's question.
You do realize that, right? You've changed subjects.
> d^3 - 1488*d^2 + 4176*d - 5
>
>directly as amazingly, in a bit of luck I guess,
>
>
> d(d^2 + 2d + 1 - 1490d + 4175) = d^3 - 1488*d^2 + 4176*d
>
>so
>
> d(d^2 + 2d + 1 - 1490d + 4175) = 5, which means
>
> d[{d + 1)^2 - 1490d + 4175] = 5
>
>which forces any root plus 1 to have the algebraic integer factors of
>any other root that is not coprime to 5.
You've proven something stronger, as you noted elsewhere. If d is any
root, and d2 and d3 are the other two roots, then d2*d3 divides
(d+1)^2.
>But looking at the solutions to
>
> d^3 - 1488*d^2 + 4176*d - 5 = 0
>
>with that information proves that at least one root must be coprime to
>5.
>
>Those solutions are
>
> {{x -> 496 + 244624/(1/2*(241976581 +
> 10683*I*Sqrt[12415]))^(1/3) +
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
>
> {x -> 496 - (122312*(1 + I*Sqrt[3]))/
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
> 1/2*(1 - I*Sqrt[3])*
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
>
> {x -> 496 - (122312*(1 - I*Sqrt[3]))/
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
> 1/2*(1 + I*Sqrt[3])*
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}}
>
>where the last two subtracted from each other give
>
> 244624/(1/2*(241976581 +
> 10683*I*Sqrt[12415]))^(1/3) +
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}
>
>times I*Sqrt[3]
To simplify, let A = (1/2*241976581 + 10683*sqrt(-12415)),
zeta = (-1+sqrt(-3))/2
zeta^2 = (-1-sqrt(-3))/2
Note that zeta+zeta^2 = -1, zeta-zeta^2 = sqrt(-3)
Then the roots are:
d1 = 496 + 244624/A^{1/3} + A^{1/3}
d2 = 496 + 2*zeta^2*122312/A^{1/3} + zeta*A^{1/3}
d3 = 496 + 2*zeta*122312/A^{1/3} + zeta^2*A^{1/3}
Subtracting the last two, we get
d2-d3 = -2(zeta-zeta^2)*122312/A^{1/3} + (zeta-zeta^2)*A^{1/3}
= -sqrt(-3)*244624/A^{1/3} + sqrt(-3)*A^{1/3}.
= (-244624/A^{1/3} + A^{1/3})sqrt(-3);
Whereas you have
(244624/A^{1/3} + A^{1/3})*sqrt(-3),
But, that's probably just a sign error.
> proving that
> 496 + 244624/(1/2*(241976581 +
> 10683*I*Sqrt[12415]))^(1/3) +
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
>
>must be coprime to 5.
Hmmm... Let's see what you are trying to do. We know that (d2+1)^2 is a
multiple of d1*d3; we know that (d3+1)^2 is a multiple of d1*d2.
So if we assume that d1 is not a unit, then d1 divides each, so
sqrt(d1) divides (d2+1) and divides (d3+1); so it divides
(d2+1)-(d3+1)=d2-d3.
Okay, so sqrt(d1) must divide the number above. So d1 must divide the
square of the number above.
But what makes you so sure that the number above is coprime to 5? We
do not even know that it is a real number (since A is a complex
number, so we need to take the cubic root of a complex number).
Please prove that
(-244624/A^{1/3} + A^{1/3})
is coprime to 5...
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Brian Quincy Hutchings
herein?
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03042...@posting.google.com>...
> We will work in the number field defined by the polynomial
>
> y^12 - 226446*y^10 + 21142596231*y^8 - 1041903035971246*y^6
> + 28575240325442631201*y^4 - 413298187541059968292500*y^2
> + 2460349402142038081821721600;
>
>
> We will let alpha be a root of this polynomial, and we will define
> algebraic integers d1, d2, d3, n1, n2, n3 in terms of alpha.
> (Taking alpha to be the real root that is approximately
> 154.451496990997802982236692688237765859998280922
> gives the numerical examples I gave earlier.
>
>
>
> MAGMA CODE:
> That proves that
>
> d1^6 - 1488*d1^4 + 4176*d1^2 - 5;
>
> has unit algebraic integer roots.
--Dec.2000 'WAND' Chairman Paul O'Neill, reelected
to Board. Newsish?
http://www.rand.org/publications/randreview/issues/rr.12.00/
http://members.tripod.com/~american_almanac
graham
>what is "MAGMA CODE," and is everyone using it,
>herein?
Magma is a computer algebra system developped at the University
of Sydney, Australia. A URL for it is
http://magma.maths.usyd.edu.au/magma/
graham
Randy Poe
> what is "MAGMA CODE," and is everyone using it,
> herein?
Apparently it's a symbolic calculator that lets you do these
abstract algebraic calculations, like finding splitting fields
of polynomials, that people need to find M-M factorizations.
JSH keeps making new threads, but if you can track back to where
Dot first posted this result, she included links on how to download
an algebra calculator so that one can verify the results
for oneself.
- Randy
Dik T. Winter
> Your question is answered rather easily by considering the polynomial
> d^3 - 1488*d^2 + 4176*d - 5
> directly as amazingly, in a bit of luck I guess,
> Those solutions are
> {{x -> 496 + 244624/(1/2*(241976581 +
> 10683*I*Sqrt[12415]))^(1/3) +
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
> {x -> 496 - (122312*(1 + I*Sqrt[3]))/
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
> 1/2*(1 - I*Sqrt[3])*
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)},
> {x -> 496 - (122312*(1 - I*Sqrt[3]))/
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3) -
> 1/2*(1 + I*Sqrt[3])*
> (1/2*(241976581 + 10683*I*Sqrt[12415]))^(1/3)}}
I think those three roots can be written as follows:
d1 = p.q
d2 = p.r
d3 = q.r
with p, q and r pairwise coprime algebraic integers... But I am still
pondering.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
James Harris
<deleted>
>
> Math is HARD because it depends on a belief in logic. And lots of
> people just don't believe in logic, as I've discovered.
>
> Your question is answered rather easily by considering the polynomial
>
> d^3 - 1488*d^2 + 4176*d - 5
>
> directly as amazingly, in a bit of luck I guess,
Nope. My attempt at finding an alternate demonstration despite the
inability of anyone to find errors in my proof, failed again.
This time I subtracted wrong.
That's wrong.
> Note that I've been forced to give demonstrations such as this for a
> PROOF that I've had for about a year now.
>
> I think it unfair and unfortunate that mathematics itself wasn't
> accepted, and I fear that even a secondary demonstration like this one
> may not sway mathematicians.
It IS unfair and unfortunate.
> I did my best. I found a short proof of Fermat's Last Theorem, and
> then people wanted more.
>
> Now I've given more, and I don't know if it will be enough.
Well the more was wrong. Oh well, it happens.
James Harris
Brian Quincy Hutchings
this is a subtle admission that
that preceding the "more" may also not be 110% correct.
eh?... I couldn't even realize,
from all of the foregoing claptrap,
taht your method was actually one of contradiction,
til you'd stated it, recently ...
which indicates some subtlety of thought, at any rate. however,
it has been shown that such inductive proofs are "one-to-one"
with deductive proofs (at least, it made sense to me,
following the 2-and-ahalf-page proof).
it should be evident from Fernat's inductive proof
of the case for N=4, that that method "plus manipulation
of real algebra" is not going to do it.
I'm not sure, though, if
you implicitly use the complex field,
as Fermat must have done, since Magidin uses it
in his back-and-forth with you.
while others may enjoy it,
I pray for a quick ending to this hilarity!
jst...@msn.com (James Harris) wrote in message news:<3c65f87.03050...@posting.google.com>...
> > I did my best. I found a short proof of Fermat's Last Theorem, and
> > then people wanted more.
> >
> > Now I've given more, and I don't know if it will be enough.
>
> Well the more was wrong. Oh well, it happens.
--les ducs d'Enron!
http://www.tarpley.net
Wayne Brown
>> I did my best. I found a short proof of Fermat's Last Theorem, and
>> then people wanted more.
>>
>> Now I've given more, and I don't know if it will be enough.
> Well the more was wrong. Oh well, it happens.
Yes, it happens. It happens every time you make another effort. You're a
failure, James. Everyone but you knows it -- and I think you know it too,
deep down inside somewhere.
--
Wayne Brown | "When your tail's in a crack, you improvise
fwb...@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"