Date: Nov 28, 2012 2:00 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem
On Wednesday, November 28, 2012 12:26:12 AM UTC+2, Pubkeybreaker wrote:

> On Nov 27, 2:58 pm, quasi <qu...@null.set> wrote:

>

> > John Jens wrote:

>

> > >quasi wrote:

>

> > >>John Jens wrote:

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> >

>

> > >>>http://primemath.wordpress.com/

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> >

>

> > >>Copying part of the text from the link above (enough to

>

> > >>expose the error in Jens' reasoning) ...

>

> >

>

> > >>>Fermat?s little theorem states that if p is a prime number,

>

> > >>>then for any integer a, the number a^p is an integer multiple

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> > >>>of p.

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> >

>

> > >>> a^p = a(mod p)

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> >

>

> > >> Yes, but note that a^p = a (mod p) does not imply 0 <= a < p.

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> >

>

> > >>>Assume that a,b,c naturals and p prime and

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> >

>

> > >>> 0 < a <= b < c < p

>

> >

>

> > >>> ...

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> >

>

> > >>>So we can?t find naturals 0 < a <= b < c < p with p prime to

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> > >>>satisfy a^p + b^p = c^p.

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> >

>

> > >> Sure, but that doesn't even come close to proving Fermat's

>

> > >> Last Theorem. All you've proved is the trivial result that if

>

> > >> a,b,c are positive integers with p prime such that

>

> > >> a^p + b^p = c^p then c >= p.

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> >

>

> > >"Assume that a , b , c naturals and p prime and 0<a=b<c<p"

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> >

>

> > Yes, you can assume anything you want, but then any conclusion

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> > is conditional on that assumption.

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> >

>

> > Without loss of generality, you can assume

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> >

>

> > 0 < a <= b < c

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> >

>

> > but how do you justify the inequality c < p?

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> >

>

> > Of course you can take 2 cases:

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> >

>

> > (1) c < p

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> >

>

> > (2) C >= p

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> >

>

> > The case you analyzed is the case c < p (the trivial case),

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> > and you never even considered the other case. Thus, you

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> > did not actually prove Fermat's Last Theorem.

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>

>

>

>

> He might also want to check his proof when p = 2.....

From the condition 0 < a <= b < c < p , p must be bigger then 2 because don't exist minimum two numbers (a <= b, c) between 0 and 2.