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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 7, 2013 1:22 AM
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On Sunday, January 6, 2013 9:05:08 PM UTC+2, M_Klemm wrote:
> You should give a reason why you assume a < p.
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> For p = 2 you have 3^2 + 4^2 = 5^2, and then by little Fermat 3 + 4
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> congruent 5 (mod 2)
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> but not 3 < 2.
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> Regards
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> Michael
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> wrote in message
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> news:696982e8-3228-48d0-b483-9cc3acb97341@googlegroups.com...
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> > On Sunday, January 6, 2013 10:56:35 AM UTC+2, M_Klemm wrote:
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> >> Hello,
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> >>
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> >>
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> >> consider the case p =3, proved by Euler. Then you see that the assumption
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> >> a
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> >> < p in line 4 is not at all justified.
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> >>
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> >> Regards
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> >> Michael
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> >
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> > I'm sorry but I don't understand what are you trying to say.
The reason is to prove FLT .
Let's split in three steps :
Step 1--> prove a^p + b^p != c^p with a < p ,a,b,c, naturals
Step 2--> extend to rationals , still a < p
Step 3--> pick A >= p, assume A^p + b^p = c^p and scaling down to A/k < p ,k rational -->contradiction to step 2
If a + b ? c>0 because 0<a?b<c implies b ? c < 0 ,
0 ? a + b ? c < a < p
then a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a < p implies p > 2
Still a + b ? c>0 ,a + b - c must be at least 1 and with p=2 we can't find natural a between a + b - c and
p there's no Step 1 for p = 2
(If a + b ? c ? 0 we have a + b ? c
(a+b)^p?c^p and using binomial theorem
a^p+b^p<(a+b)^p=a^p+?+b^p?c^p)
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