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FLT - The True Nature of this Problem Revealed
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master1729  
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 More options Nov 3, 11:10 pm
Newsgroups: sci.math
From: master1729 <tommy1...@gmail.com>
Date: Tue, 03 Nov 2009 18:10:03 EST
Local: Tues, Nov 3 2009 11:10 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
Phil Jackson wrote :

> This paper has been shown to several mathematicians
> including a highly respected researcher. They have
> not been able to break this proof but they still
> don't want to accept it, for reasons that are nothing
> to do with the logic.

> I believe that the simple logic cannot be faulted but
> want to test that against a wider audience.

> I'm not an academic and some of the terminology may
> be used in a context that is in conflict with its
> traditional usage but it should be simple enough for
> anyone to quickly grasp.

not even close to a proof.

jsh did better !

im trying to find a short proof myself.

i once tried to prove a short proof doest exist.

but failed and at the moment try to find a short proof myself.

also because i once tried to prove a short proof of RH doesnt exist.

and now i now it actually does exist. ( " relatively short " not as short as your doc of course :p  lol sorry )

that drives me back to FLT.

but FLT is one though son of a bitch...

anyways , its clear you need to practice your number theory skills.

regards

tommy1729


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Phil Jackson  
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 More options Nov 4, 12:15 am
Newsgroups: sci.math
From: Phil Jackson <sapi...@clear.net.nz>
Date: Tue, 03 Nov 2009 19:15:32 EST
Local: Wed, Nov 4 2009 12:15 am
Subject: Re: FLT - The True Nature of this Problem Revealed
Hi Tommy

Can you tell me where the flaw is, in arriving at the statement that Z = Y+X1, where X1 is a factor of X^N?

Regards

Phil


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Achava Nakhash, the Loving Snake  
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 More options Nov 4, 5:50 am
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Tue, 3 Nov 2009 21:50:07 -0800 (PST)
Local: Wed, Nov 4 2009 5:50 am
Subject: Re: FLT - The True Nature of this Problem Revealed
On Nov 3, 4:15 pm, Phil Jackson <sapi...@clear.net.nz> wrote:

> Hi Tommy

> Can you tell me where the flaw is, in arriving at the statement that Z = Y+X1, where X1 is a factor of X^N?

> Regards

> Phil

Hi Phil,

I will speak to the flaws in your presentation.  Since this thread
doesn't give a clue as to what you said and where you said it, I had
to hunt before I could find it.  It would be easier for the group and
help you get replies if you would give a link or at least a clue as to
where to find your work.  Of course the responsibility was really
master1729's.

Anyway, let's start with what you said about the case N = 2.  Here is
what you said:

Start Quote
What happens when N=2?

When N=2, Z^2 – Y^2 = X^2 and this can be broken down into (Z-Y)(Z+Y)
so that X^2 is the product of (Z-Y) or X1, and (Z+Y) or X2. Solutions
can be found for any situation where X1 is an odd power of X and X2 is
a higher odd power of X multiplied by 1 or another number raised to
the same power. Therefore it is clear that there must be an infinite
number of solutions for N=2.
End Quote

Really?  Why is that clear?  You are claiming that it is clear (=
obvious) that we can always find Y and Z such that Y-Z is an odd power
of X and Y+Z ... as you said above.  First of all, this is simply
wrong for then their product could be larger than X^2 and larger is
not equal to.  However, they only need to be an odd power of anything
and so on as you say without reference to the X.  But you need to
prove that you can always find that and it is really far from obvious
to me.  I only know that this equation, X^2 + Y^2 = Z^2 has infinitely
many solution because I have read an actual proof that doesn't simply
leave out the most important step.

Now to your absurd claim about N larger than 2.  It is hard for me to
point to a specific error, because it is completely incoherent.  Write
it coherently and I will be glad to review and tell you where your
errors are.

Notes on Style:
1) You spend a long time justifying the obvious fact that if Z^3 - Y^3
= X^3 then X^3 is divisible by Z - Y.  I say it is obvious because I
remember the factorization

Z^2 - X^3 = (Z - Y)*(Z^2 + ZY + Y^2)

which I learned in a Junior High School algebra course and still
remember.  You never actually present the analogue for N larger 3, but
the same general idea works for any N.

2) You spend a lot of time making the point that X^N + Y^N = Z^3 is
logically equivalent to
Z^N - Y^N = X^N.  Yes, you can subtract the same things from both
sides of an equation and the equality still holds.   You can add the
same thing too, so the resulting equation is logicallyl equivalent.
Duh.  Just state things like this.  They need no justification.

3)  In you proof section, you say
Start Quote
There is evidence for a relationship between two numbers where one is
subtracted from the other when they have the same power.
End Quote

There is rarely a place for saying that there is evidence for
something in a proof as part of the proof.  Your style will be
improved if you just stick to the facts.

4) Similarly to point 3), your proof section is very chatty, and that,
combined with the fact that a lot of the chat does not make a lot of
sense, makes the part of your "proof" that do make sense harder to
follow, probably for you as well as for the reader.

5) The nonsensical philosphical ramblings in your introduction mostly
serve to drive away mathematicians who immediately pick up your vast
ignorance of our subject and who do not have the patience to read what
is already clear will do what you say it will do.

That is enough for now.  I only read the thing to try to give you some
pointers as to what was specifically wrong, but it was hard to find
much that was coherent enough to comment on.

I appreciate that you have a love of mathematics, but please to try to
learn some before you attempt something as grand as Fermat's Last
Theorem, or any thing much else for that matter.

Regards,
Achava


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master1729  
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 More options Nov 4, 2:23 pm
Newsgroups: sci.math
From: master1729 <tommy1...@gmail.com>
Date: Wed, 04 Nov 2009 09:23:53 EST
Local: Wed, Nov 4 2009 2:23 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
the loving snake wrote :

responsibility denied.

i had nothing meaningfull to link too and i am not responsible for wrong FLT proofs by others nor for not quoting nonsensical wrong FLT proofs.

the doc was simply available in the first post of the thread.

no problems for viewing.

agreed.

> 5) The nonsensical philosphical ramblings in your
> introduction mostly
> serve to drive away mathematicians who immediately
> pick up your vast
> ignorance of our subject and who do not have the
> patience to read what
> is already clear will do what you say it will do.

which is why i did not quote a part of his " paper ".

its very noble that you spend time trying to help this person.

however the ramblings give a really bad impression.

i hope this guy is not the new musatov.

otherwise you will regret you did the effort.

basicly he just showed x^p + y^p = z^p => x + y = z mod p

or even less.

which already follows from trivial number theory such as fermats little , newtons binomium , pascal triangle etc

> That is enough for now.  I only read the thing to try
> to give you some
> pointers as to what was specifically wrong, but it
> was hard to find
> much that was coherent enough to comment on.

> I appreciate that you have a love of mathematics, but
> please to try to
> learn some before you attempt something as grand as
> Fermat's Last
> Theorem, or any thing much else for that matter.

love for math is cool.

> Regards,
> Achava

from x^p + y^p = z^p => x + y = z mod p

its clear z must be >= p  ( if x,y,z > 0 )

( maybe phil can use that knowledge )

i wonder if achava the loving snake can prove :

z must be >= 12 p  and x must be >= 3 p

in a simple way.

no hostility intended ( despite received ? )

regards

tommy1729


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Achava Nakhash, the Loving Snake  
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 More options Nov 4, 4:51 pm
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Wed, 4 Nov 2009 08:51:04 -0800 (PST)
Local: Wed, Nov 4 2009 4:51 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
On Nov 4, 6:23 am, master1729 <tommy1...@gmail.com> wrote:

No hostility intended to you, Tommy.  I read this group on google and
the original post did not show up.  I will take your challenge when I
get a chance, which could be a while, since I have several things
stacked up.

Regards,
Achava


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Phil Jackson  
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 More options Nov 4, 9:05 pm
Newsgroups: sci.math
From: Phil Jackson <sapi...@clear.net.nz>
Date: Wed, 04 Nov 2009 16:05:25 EST
Local: Wed, Nov 4 2009 9:05 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
Hi Achava

I have addressed all your points and shown in the updated paper I posted that Z=Y+X1, where X1 is a factor of X^N

I would be very interested if you prove that the above is false? I could elaborate more on this relationship if you are unable to grasp it -  the earlier mathematicians I showed this to understood the relationship clearly although they were most unhappy with the outcome.

I was also interested in the following;
from x^p + y^p = z^p => x + y = z mod p
its clear z must be >= p  ( if x,y,z > 0 )

I presume that P is greater than 2 as there are numerous counter examples for N=2

e.g. 25^2 + 60^2 = 65^2 but 25+60 <> 65 mod 2

Regards

Phil


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Gerry Myerson  
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 More options Nov 4, 10:44 pm
Newsgroups: sci.math
From: Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
Date: Thu, 05 Nov 2009 09:44:01 +1100
Local: Wed, Nov 4 2009 10:44 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
In article
<616504531.11852.1257368755646.JavaMail.r...@gallium.mathforum.org>,
 Phil Jackson <sapi...@clear.net.nz> wrote:

> I was also interested in the following;
> from x^p + y^p = z^p => x + y = z mod p
> its clear z must be >= p  ( if x,y,z > 0 )

> I presume that P is greater than 2 as there are numerous counter examples for
> N=2

> e.g. 25^2 + 60^2 = 65^2 but 25+60 <> 65 mod 2

If you are saying 25 + 60 is not congruent to 65 (modulo 2)
then you might want to reconsider what you are doing.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)


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Achava Nakhash, the Loving Snake  
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 More options Nov 4, 11:37 pm
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Wed, 4 Nov 2009 15:37:03 -0800 (PST)
Local: Wed, Nov 4 2009 11:37 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
On Nov 4, 6:23 am, master1729 <tommy1...@gmail.com> wrote:

Tommy,

I havd now learned why it appeared that you neglected to post the
document you were commenting on.  The original post was on MathForum.
I read sci.math on google groups.  Apparently posts from MathForum
with attachments are not visible to those of us who use google
groups.  I saw no original post and no document.  I eventually found
the post on Math Forum, a site with which I was not previously
familiar except for the familiar ranting about Math Forum morons which
I always ignore.  Anyway, it is probably a good idea to repost what
you are responding to if you know it came from Math Forum.  Some
people seem to know this, although I do not know how.

Regards,
Achava


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Phil Jackson  
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 More options Nov 5, 1:55 am
Newsgroups: sci.math
From: Phil Jackson <sapi...@clear.net.nz>
Date: Wed, 04 Nov 2009 20:55:55 EST
Local: Thurs, Nov 5 2009 1:55 am
Subject: Re: FLT - The True Nature of this Problem Revealed
Hi Achava

I have a question for you.

You said that you were aware for N=3, that the difference between Z^N and Y^N was divisible by the difference between Z and Y (or X1).

Knowing that and that trying to turn Z = Y + X1 into Z^3 = Y^3 + X^3 is an impossible task, why is it that no one has realised that at least for N=3, there was a very simple reason why Z^3 = Y^3+X^3 had no solutions?

If is very clear to anyone that Z-Y has to be a factor of X^3 for there to be a solution, and therefore why is it that no one spotted this simple proof? You overlooked it yourself.


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Achava Nakhash, the Loving Snake  
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 More options Nov 5, 4:04 am
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Wed, 4 Nov 2009 20:04:16 -0800 (PST)
Local: Thurs, Nov 5 2009 4:04 am
Subject: Re: FLT - The True Nature of this Problem Revealed
On Nov 4, 5:55 pm, Phil Jackson <sapi...@clear.net.nz> wrote:

> Hi Achava

> I have a question for you.

> You said that you were aware for N=3, that the difference between Z^N and Y^N was divisible by the difference between Z and Y (or X1).

> Knowing that and that trying to turn Z = Y + X1 into Z^3 = Y^3 + X^3 is an impossible task, why is it that no one has realised that at least for N=3, there was a very simple reason why Z^3 = Y^3+X^3 had no solutions?

> If is very clear to anyone that Z-Y has to be a factor of X^3 for there to be a solution, and therefore why is it that no one spotted this simple proof? You overlooked it yourself.

Phil,

It is obvious that if Z^3 = Y^3 + X^3, then Z - Y divides X^3.
Consider it accempted by everyone.  Nobody has missed it.  So what?
How does that morph into a proof?

Regards,
Achava


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Phil Jackson  
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 More options Nov 5, 8:56 pm
Newsgroups: sci.math
From: Phil Jackson <sapi...@clear.net.nz>
Date: Thu, 05 Nov 2009 15:56:48 EST
Local: Thurs, Nov 5 2009 8:56 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
Hi Achava

If Z - Y = X1 (a factor of X^3) then how do you turn
Z = Y + X1 into Z^3 = Y^3 + X^3?

If I do the obvious and multiply Z by Z^2 I get Z^3, but I also get (Z^2 x Y) + (Z^2 x X1) which is not Y^3 + X^3.

I can also multiply Y by Y2 and get (Y^2 x Z) - (Y^2 - X1) which is not Z^3 - X^3.

That also fails! Like a result that produces 1=0, this is a nonsensical contradiction. Z = Y + X1 is verifiable but you end up trying to compare it to something it clearly cannot be.

If you disagree with that (and that would courageous), then prove me wrong? Find a counter-example and I will believe you.


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Peter  
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 More options Nov 5, 9:11 pm
Newsgroups: sci.math
From: Peter <perumbra...@gmail.com>
Date: Thu, 05 Nov 2009 16:11:32 EST
Local: Thurs, Nov 5 2009 9:11 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
Dear Philip

I've looked at your paper and would like to make some suggestions. Your second version is an improvement but you should really remove all the ideas about complexity. As Achava pointed out, these types of comments are immediately going to put someone on the offensive when you actually raise some valid points. They will be lost amongst the criticism.

You're obviously not a trained mathematician but unlike others in this topic I can also follow your logic and although there is some ambiguity I can see your point reasonably clearly.

I think that you've come up against some long-established axioms that your paper appears to contradict and mathematicians like to be safe and go with what they already accept and have spent a considerable amount of time building other concepts on top of.

Yet your first step of proving the relationship is sound and accepted.

Your second step of then showing how this creates an impossible result is interesting and I cannot fault it or find any logic that disproves it.

I personally think that your are possibly right but I will wait to see if anyone can contradict this step.


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master1729  
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 More options Nov 5, 10:09 pm
Newsgroups: sci.math
From: master1729 <tommy1...@gmail.com>
Date: Thu, 05 Nov 2009 17:09:52 EST
Local: Thurs, Nov 5 2009 10:09 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
Phil Jackson wrote :

> Hi Achava

> If Z - Y = X1 (a factor of X^3) then how do you turn

> Z = Y + X1 into Z^3 = Y^3 + X^3?

z = y + x1 => z^3 = x1^3 + 3y(x1)^2 + 3(y)^2 x1 + y^3

together with z^3 = y^3 + x^3

gives z^3 - y^3 - x^3 = 0 = (y + x1)^3 - y^3 - x^3

thus 0 = x1^3 + 3y(x1)^2 + 3(y)^2 x1 - x^3. ( if x1 is a factor of x^3 , read as ' condition ' )

so how does 0 = x1^3 + 3y(x1)^2 + 3(y)^2 x1 - x^3 help in proving FLT according to you ???

how does 0 = x1^3 + 3y(x1)^2 + 3(y)^2 x1 - x^3 imply that Z^3 =/= Y^3 + X^3 ??

> If you disagree with that (and that would
> courageous), then prove me wrong? Find a
> counter-example and I will believe you.

i am very courageous !!

tommy1729


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Achava Nakhash, the Loving Snake  
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 More options Nov 6, 6:49 pm
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Fri, 6 Nov 2009 10:49:21 -0800 (PST)
Local: Fri, Nov 6 2009 6:49 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
On Nov 5, 12:56 pm, Phil Jackson <sapi...@clear.net.nz> wrote:

> HiAchava

> If Z - Y = X1 (a factor of X^3) then how do you turn
> Z = Y + X1 into Z^3 = Y^3 + X^3?

> If I do the obvious and multiply Z by Z^2 I get Z^3, but I also get (Z^2 x Y) + (Z^2 x X1) which is not Y^3 + X^3.

> I can also multiply Y by Y2 and get (Y^2 x Z) - (Y^2 - X1) which is not Z^3 - X^3.

> That also fails! Like a result that produces 1=0, this is a nonsensical contradiction. Z = Y + X1 is verifiable but you end up trying to compare it to something it clearly cannot be.

> If you disagree with that (and that would courageous), then prove me wrong? Find a counter-example and I will believe you.

Hi Phil,

You start with a demonstrable fact, namely that if (X, Y, Z) is a
solution of the cubic Fermat equation, then  Z - Y = X1 is a factor of
X^3, and then you ask how that can be turned into
X^3 + Y^3 = Z^3.  You then try two approaches that fail, and
apparently draw the conclusion that there be no solutions from the
fact that you made two tries and both of them didn't work.  The logic
is so weird that it is unclear to me how to refute it in a way that
would be meaningful to you, but of course I will try.

First of all, it is completely irrelevant how many failed attempts you
make.  Your failure to do something could be caused by its
impossibility, but it could also be explained by your personal
inabilities.  It is not a proof of anything.  To prove that there are
no solutions to
X^3 + Y^3 = Z^3 you have develop a contradiction based on that fact
alone.  You are certainly allowed to include deductions sucy as that Z
- Y must be a factor of X^3.  But you must show that any POSSIBLE
attempt fails, not just the ones you tried.  Or don't even worry about
attempts to make the equation happen.  Just show that it can't be true
because it implies a contradiction.

Although I have certainly never seen such an elementary approach
succeed, even for the cubic form of the Fermat equation, I would
certainly not rule out its possibility.  The likelihood that it could
be done for the cubic Fermat is even higher.  I recently read a paper
describing how to solve a certain quartic diophantine equation which
Richard Bumby had asked the author to try and prove by what the next
author considered the most natural method.  For this guy, the most
natural methods were to use somewhat sophisticated algebraic number
theory.  They were for me too, but I had forgotton so much of what I
used to know that I could not do it that way, so I found a way that
involved only high school algebra and the unique factorization theorem
of integers.  I intend to post that at some point, and maybe I will
get around to this weekend.

I see from Peter's post below that your updated paper is accessible to
at least someone.  I looked for it a couple of days ago, but could not
find it.  I read sci.math on Google Groups, and it does not seem to
pick up attachments from Math Forum, so that could be the problem.
Maybe you could give us all the URL for your updated paper on your
next post in this thread.

Regards,
Achava


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Achava Nakhash, the Loving Snake  
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 More options Nov 8, 6:13 am
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Sun, 08 Nov 2009 01:13:56 EST
Local: Sun, Nov 8 2009 6:13 am
Subject: Re: FLT - The True Nature of this Problem Revealed
Hi Phil,

I didn't see this post (your reply to mine - I am not sure why that is showing up below, but I haven't used Math Forum before.  I didn't see your reply because posts done through Math Forum that have attachments are for some reason not visible through google groups which is what I have always used before. Anyway -

You say that the case for N = 2 can be rewritten as

(Z-X)(Z+X) = Y^2 or X1*X2 = Y^2

Then you go on to say

"Solutions can be found for any situation where X1 is an odd power of X and X2 is a higher odd power of X multiplied by 1 or another number raised to the same power. Therefore it is clear that there must be an infinite number of solutions for N=2."

The bit about a higher odd power of X makes no sense, but it is true that if X2 is equal ot X1 or on odd power of X1 or many other possible things, then the product is a square.  The point is that you have to choose both  Z and X so that this is true, and you give no clue how this can be done for even a single triple (X, Y, Z) much less infinitely many.  You have proved nothing.  Please think carefully about what you are saying and let us know when you have a version that actually proves something.  Vagueness does not pass muster.  YOu must either provide all the details or else provide enough details that we can readily fill in the rest.

Regards,
Achava


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master1729  
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 More options Nov 8, 2:55 pm
Newsgroups: sci.math
From: master1729 <tommy1...@gmail.com>
Date: Sun, 08 Nov 2009 09:55:12 EST
Local: Sun, Nov 8 2009 2:55 pm
Subject: Re: FLT - The True Nature of this Problem Revealed
the loving snake wrote :

> On Nov 4, 6:23 am, master1729 <tommy1...@gmail.com>
> wrote:

(snip)

still waiting for achava the loving snake taking up the challange ...

regards
tommy1729


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