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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 7, 2013 10:37 AM
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On Monday, January 7, 2013 3:18:29 PM UTC+2, M_Klemm wrote:
> "John Jens" wrote
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> > The reason is to prove FLT .
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> > Let's split in three steps :
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> > Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c, naturals
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> > Step 2--> extend to rationals , still a < p
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> > Step 3--> pick A >= p, assume A^p + b^p = c^p and scaling down to A/k < p
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> > ,k rational -->contradiction to step 2
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> > If a + b ? c>0 because 0<a?b<c implies b ? c < 0 ,
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> > 0 ? a + b ? c < a < p
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> > then a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a <
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> > p implies p > 2 ...
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> > .... and using binomial theorem
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> Is this intended to be a proof of step 1?
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> If yes, it is essentially correct, because a^p + b^p = c^p together with a <
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> p implies
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> a^p + b^p <= a + b^p < p + b^p < (b+1)^p <= c^p, a contradiction.
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> The inequality (b+1)^p <= c^p is however not necessaryly true for rational b
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> and c with b < c.
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> Regards
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> Michael
Yes it's only for step 1 when a,b,c naturals.
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