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Math Forum / Mathematics / General Topics / September 2004



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Fermat's last theorem (New)

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ben ito - 30 Sep 2004 07:52 GMT
Fermat's Last Theorem
Ben T. Ito
9-29-04

I will solve Fermat's last theorem.

1. Introduction

Fermat implied that there was a simple proof that for X^n + Y^n = Z^n
only n=2 forms integer solutions that describe lengths X, Y and Z.

2. Proof

Using the lengths X=3 and Y=4, when n is at infinity and approaches
two, the length Z approaches five.  At n=2, Z=5 which forms an
integer solution that forms a right triangle.

An infinite number of proportional right triangles solutions can be
formed using,

X=3m, Y=4m, and Z=5m,     equ 1

where m=1,2,3,4............ .   An infinite number of integer
solutions are formed when the right triangle are formed using
proportional right triangles, of equation l, when n=2 ; therefore,
only right triangles, when n=2, form integer solutions.

3. Conclusion

I have formed an infinite number of solutions, using proportional
right triangles, that only form integer solutions when n=2 which
completes the proof of Fermat's last theorem.

Sonya Stewart.

--------------------------------------------------------------------------------
Jeroen Boschma - 30 Sep 2004 11:06 GMT
> Fermat's Last Theorem
> Ben T. Ito
[quoted text clipped - 37 lines]
>     ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
> ----------------------------------------------------------

Instead of blindly restating your 'proof' over and over again, (JSH tactics), it may be nice to
answer questions also. I'll restate mine:

Can you give an argument, based on your own proof, why there is not a certain Z and n>2 such that,
for example,  17^n + 53^n = Z^n ???

Your proof sounds something like: because I can form an infinite number of solutions for n=2, there
just cannot be other solutions. Strange...

  Jeroen
Jim Ferry - 30 Sep 2004 16:46 GMT
> > I will solve Fermat's last theorem.

> > I have formed an infinite number of solutions, using proportional
> > right triangles, that only form integer solutions when n=2 which
> > completes the proof of Fermat's last theorem.

> Your proof sounds something like: because I can form an infinite number
> of solutions for n=2, there just cannot be other solutions. Strange...

Oh, I think it's quite nice.  For example, if you have 4 objects in a set,
and you can show that 4 objects in that set are elephants, then it follows
that every member of S is an elephant.

Indeed, for any finite number N, if a set S has N members, and if N members
of S have a certain property, then all members of S have that property.

Now using the fact that infinity is a finite number -- oh, I know this may
be counterintuitive, but that's just because crafty Cantorians *named* things
to make it seem like infinity is infinite -- so, again, using the fact that
infinity is a finite number, we can prove a variety of things.

For example, let S be the set of solutions of X^n + Y^n = Z^n.  An infinite
number of them occur for n=1, so all solutions occur for n=1.  In particular,
none occur for n>2.  Or for n=2, for that matter.  And this holds whether
you're working in the ring of integers, or some neo-Harrisian object ring.

On the other hand, because there are an infinite number of solutions to
X^n + Y^n = Z^n, all 4-tuples (X,Y,Z,n) are solutions.  Etc.

The thing I like about this method is that it gives Cantor cranks something
to talk about with Fermat cranks.  And cranks are usually such solitary
creatures.
Dimitris - 30 Sep 2004 14:29 GMT
> I will solve Fermat's last theorem.

Cool. Go on to prove Goldbach's conjecture as well. Since there are infinite
odd primes, you can add 3 to each of them, thus producing an infinity of even
numbers which are a sum of two primes. QED.

Dimitris
Geert van der Wulp - 30 Sep 2004 15:05 GMT
On 30 Sep 2004, George wrote:
>benito20044@yahoo-dot-com.no-spam.invalid (ben ito) wrote in message news:<415bad40$1_2@127.0.0.1>...
>> Fermat's Last Theorem
[quoted text clipped - 35 lines]
> I don't see in any case that from this we can conclude that only when
>n=2 we have only integer solutions.

I think that the point is that none of the solutions for X, Y and Z which "work" for n = 2 will work for n > 2. But even if this is the case, what about all other possible combinations of X, Y and Z? Why would we only want to a Pythagorean triple?

>> 3. Conclusion
>>
[quoted text clipped - 14 lines]
>>     ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
>> ----------------------------------------------------------        

Geert
 
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