|
|
Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Nov 27, 2012 5:26 PM
|
|
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.....
|
|
