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Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 7, 2013 8:18 AM
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"John Jens" wrote
> 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 ...
> .... and using binomial theorem
Is this intended to be a proof of step 1?
If yes, it is essentially correct, because a^p + b^p = c^p together with a <
p implies
a^p + b^p <= a + b^p < p + b^p < (b+1)^p <= c^p, a contradiction.
The inequality (b+1)^p <= c^p is however not necessaryly true for rational b
and c with b < c.
Regards
Michael
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