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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Nov 28, 2012 1:49 AM
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On Tuesday, November 27, 2012 9:58:32 PM UTC+2, quasi wrote:
> John Jens wrote:
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> >quasi wrote:
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> >>John Jens wrote:
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> >>
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> >>>http://primemath.wordpress.com/
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> >>
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> >>Copying part of the text from the link above (enough to
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> >>expose the error in Jens' reasoning) ...
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> >>
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> >>>Fermat?s little theorem states that if p is a prime number,
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> >>>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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> >>
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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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> >>>
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> >>> 0 < a <= b < c < p
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> >>>
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> >>>
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> >>> ...
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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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> >>
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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
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>
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> 0 < a <= b < c
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>
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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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>
>
> 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)...
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