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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Dec 1, 2012 12:20 AM
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On Tuesday, November 27, 2012 10:49:45 PM UTC-8, John Jens wrote:
> On Tuesday, November 27, 2012 9:58:32 PM UTC+2, quasi wrote:
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> > John Jens wrote:
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> > >quasi wrote:
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> > >>John Jens wrote:
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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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> > >>> 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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> > >> 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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> > >"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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> > quasi
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> If a > c and/or b > c it's obvious that a^p + b^p > c^p.
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> 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
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> (a + mp)^p?(a + mp)(modp)=a + mp + kp=a + p(m+k)...
You don't know when to stop, do you?
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