quasi
Posts:
12,042
Registered:
7/15/05


Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Nov 27, 2012 2:58 PM


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

