Fermats Last Theorem

To Prove that there is no Z, where Z is a positive whole number such that Z^n=Y^n +X^n and X,Y,n are positive whole numbers and n>2.

The structure of the proof is to prove that there is no whole numbers on the line segment from Y^2 to (X^2+Y^2)., if you assume that the equality Z^n=X^n+Y^n holds. A contradiction will be proved for every whole number on that line segment, thus proving that the equation is wrong for every Z.

First, the starting point of the segment (Y^2) will be shown to be less than Z^2 in every instance of Z.  Z^n will shown to be larger than Y^2 for  every instance of  n.

Second, there will be an examination of the line segment from Y^2 to (Y^+1)^2. It will be shown that Z^2 must be greater than the numbers Y^2 +1, Y^2 +1, 

Y^2 +3, ..... + (Y^2+1) -1

Lemma 1

This approach cannot be used when Z^2 is assumed to be (Y+1)^2, so a different approach is used when Z is assumed to be an whole number squared in this line segment.

Third, when Z^2 is assumed to be less than or equal equal to (Y+1)^2, there will be shown to be a  cantridiction with the assumption that X,Y are whole numbers. The implication is that Z^2 is greater than the square of that whole number.

Lemma 2

Forth We return to the first argument, because Z^2 is now > (Y + 1)^2  and look at the line segment from (Y+1)^2 to (Y+2)^2 -1. Z^2 is again shown that it cannot be  any whole number  in this set, and must be >  (Y +2)^2 -1.

Lemma 3

Fifth Z^2 is now assumed to be (Y+2)^2 This  is shown to be in contradiction with assumption  that  X,Y are whole numbers, and that Z^2 must be > (Y+2)^2

Six. The pattern should now be clear that one argument advances Z^2 from being greater than (Y +M )^2, where M is a whole number to being greater than  

(Y+M+1)^2 -1. The second argument is used when Z^2 = (Y + M)^2, and this advances Z^2  past the squared whole number, and sets up another round of arguments.

Lemma 4

Seven Finally, the Z^2 advances until it is greater than (X^2 + Y^2) -1. Once again the first argument is used to show that the only possibility for Z^2 is equal to 

X^2+Y^2. But then this means that n=2, a contradiction with n>2.

Here are the Lemmas that are in the proof.

1^st Lemma (Prove Z^2 cannot be any whole number between Y^2 and (Y +1)^2)

Assume Z^2 = Y^2 +1, and prove contradiction. (Y + 1)^2 > Y^2 +1, because (Y+1)^2 = Y^2 + 2Y +1. If Y is whole number, Z^2 > Y^2 + 1. If M < or=" 2Y,"> Y^2 + 2Y..

2nd Lemma (  Prove that Z^2 > (Y +1)^2  )

Assume that Z^2 = (Y + 1)^2. I would like to introduce a function F such that F^m = (X^m +Y^m)^1/m  Z=(X^n +Y^n)^1/n,is a member of this function.

When Z^2 = (Y+1)^2, F^2 > (Y+1)^2. This implies there is some F^c = (Y+1)^2 where 1 < c =" (Y">

3^rd Lemma (Prove that that for any N, Z^2 > (Y+N)^2 + 2(Y + N), given that Z > Y+N  )

Then argument used in 1^st Lemma applies and Z^2  is finally > ((Y + N)^2 + 2(Y + N) )

4^th Lemma ( Prove that Z^2 > (Y+N)^2 )

Assume that Z^2 = (Y + N)^2. Then by 2^nd Lemma, Z^2 = (Y + M)^2 and again there  is a F^d such that F^d=(Y+M)^2    Again, 1 < d =" (Y">

5^th Lemma ( Prove that Z^2 is not equal to X^2 + Y^2 )

Assume Z^2 = X^2 + Y^2. Then Z^n > X^n + Y^n for every n > 2, a contradiction with original assumptions.

 

Proof

Every whole number between Y^2 to X^2 + Y^2 has been disqualified from being a answer to Z. Therefore, there is no Z that can be an answer to Z^n =X^n + Y^n