Date: Nov 28, 2012 1:49 AM
Author: John Jens
Subject: Re: From Fermat little theorem to Fermat Last Theorem
On Tuesday, November 27, 2012 9:58:32 PM UTC+2, quasi wrote:

> John Jens wrote:

>

> >quasi wrote:

>

> >>John Jens wrote:

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> >>

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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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> >>>

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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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> >>

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

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> >>>

>

> >>>

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> >>>

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

>

> >>>

>

> >>>

>

> >>> ...

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> >>>

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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

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

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

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

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> >

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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

>

>

>

> 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:

>

>

>

> (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.

>

>

>

> quasi

If a > c and/or b > c it's obvious that a^p + b^p > c^p.

We ca choose z ,y ,z > p , x <= y < z and using modulus properties a ,b ,c, a <= b < c ,a < p that x = a + mp ,y = b + np, z = c + qp with m, n ,q naturals

(a + mp)^p?(a + mp)(modp)=a + mp + kp=a + p(m+k)...