
Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Nov 28, 2012 1:49 AM


On Tuesday, November 27, 2012 9:58:32 PM UTC+2, quasi 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. > > > > quasi If a > c and/or b > c it's obvious that a^p + b^p > c^p. 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 (a + mp)^p?(a + mp)(modp)=a + mp + kp=a + p(m+k)...

