Date: Nov 27, 2012 2:58 PM
Author: quasi
Subject: Re: From Fermat little theorem to Fermat Last Theorem
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