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