```Date: Nov 27, 2012 5:26 PM
Author: Pubkeybreaker
Subject: Re: From Fermat little theorem to Fermat Last Theorem

On Nov 27, 2:58 pm, quasi <qu...@null.set> wrote:> John Jens wrote:> >quasi wrote:> >>John Jens wrote:>> >>>http://primemath.wordpress.com/>> >>Copying part of the text from the link above (enough to> >>expose the error in Jens' reasoning) ...>> >>>Fermat?s little theorem states that if p is a prime number,> >>>then for any integer a, the number a^p is an integer multiple> >>>of p.>> >>>   a^p = a(mod p)>> >> Yes, but note that a^p = a (mod p) does not imply 0 <= a < p.>> >>>Assume that a,b,c naturals and p prime and>> >>>   0 < a <= b < c < p>> >>> ...>> >>>So we can?t find naturals 0 < a <= b < c < p with p prime to> >>>satisfy a^p + b^p = c^p.>> >> Sure, but that doesn't even come close to proving Fermat's> >> Last Theorem. All you've proved is the trivial result that if> >> a,b,c are positive integers with p prime such that> >> a^p + b^p = c^p then c >= p.>> >"Assume that a , b , c naturals and p prime and 0<a=b<c<p">> Yes, you can assume anything you want, but then any conclusion> is conditional on that assumption.>> Without loss of generality, you can assume>>    0 < a <= b < c>> but how do you justify the inequality c < p?>> Of course you can take 2 cases:>>    (1) c < p>>    (2) C >= p>> The case you analyzed is the case c < p (the trivial case),> and you never even considered the other case. Thus, you> did not actually prove Fermat's Last Theorem.He might also want to check his proof when p = 2.....
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