Analysis using Diopphantine Equations

Diophantine Equations are used to remove as many variables as possible and write the remaining unknowns in terms of the other unknowns. By analyzing the remaining terms as whole numbers, we can decided whether there are infinite number of solutions or zero solutions.

Because we have the given equation X^n + Y^n = Z^n.

From this equation we can derive to associated equations

1) X + Y = c/d * Z where c/d > 1. c,d are rational also c/d < 2^1/2

2) X^2 + Y^2 = e/f * Z^2 where e/f > 1 and c/d > e/f e,f are rational


from 1) we get d*X + d*Y = C*Z

d*X = c*Z - d*Y

or finally X = c*Z/d - Y


If we substitute value for X into 2) we get

( c^2/d^2 * Z^2 - 2c/d * Z * Y + Y^2) + Y^2 = e/f * Z

rearranging

c^2/d^2 * Z^2 - 2c/d * Z * Y - e/f*Z = -2Y^2

or

Z *(c^2/d^2 * Z - 2c/d * Y - e/f) = -2Y^2

Now let Z = h *Y where 2 > h > 1 and substitute into equation

h * Y *((c^2/d^2 * h *Y) - 2c/d * Y - e/f) = -2Y^2

factor Y out of both sides of equation


and you get


h *((c^2/d^2 * h *Y) - 2c/d * Y - e/f) = -2Y

Now move -2Y to left hand side and extend h across members of brackets

c^2/d^2 * h^2 *Y - 2c/d * h * Y + 2Y - e/f * h = 0

move e/f * h to right hand side. Now left hand side has the Y come out of all terms on left hand side


(c^2/d^2 * h^2 - 2c/d * h + 2) * Y = e/f * h

or finally

Y = (e/f * h)/(c^2/d^2 * h^2 - 2 * c/d * h + 2)

We do not know what the exact value of y is, but we can determine the range

when Y is close to Z, e/f approximates 1, h approximates 1, c^2/d^2 approximates 1, but is always less than 1

In that case

Y is approximately (1 * 1)/((1 * 1^2) - 2 * 1 * 1 + 2) = 1/1 = 1, but less than 1




When Y is close to X, e/f approximates 1.3, h approximates 2, c^2/d^2 approximates 1.4

Then
Y is approximately (1.3 * 2)/(1.4^2 * 2^2 - 2 * 1.4 * 2 + 2)= 2.6/7.84 - 5.6 + 2

or Y is approximately 2.6/4.24 or approximately 0.6

These approximations could be made more defined, but the real issue has already been decided.

The range of values Y takes on is between 0.6 and 1. There is no way Y could be equal to 2, even if the numbers are refined. It should be noted that Y is the larger of the supposed values X,Y. This only leaves an integer value of less than 0 for X, which is not allowed.

This completes the proof for Fermat's Last Theorem using Diophantine Equations.

What happens when Z is presumed to be whole

Let's see when Z is presumed to be a whole number

Let's look at the section of the number line from Y^2 to (X + Y)^2

It should look like this:

------- Y^2 ------ Z^2 ----- Z^2 + 1 ----------(X^2 + Y^2)----(X + Y)^2 ----

We know there is at least one number that corresponds to Z^2 + 1 since (X^2 + Y^2) is a member of that set of whole numbers between Z^2 and (X + Y)^2. There may be many numbers,there may be only one number, but there is at least one.

Let us now look at the number line from Y^n to (X + Y)^n

Z^n
--------Y^n -------+ ------Z^n + 1 -------- (X^2 + Y^2)n/2 -----(X + Y)^n----
(X^n + Y^n)

If you inspect closely, in the first line segment Z^2 + 1 < (X^2 + Y^2) and in the

second (X^n + Y^n) < Z^n +1.

What this implies is that there is some Z^h where

where 2 < h < n, where Z^h = X^h + Y^h.

We also know h is 'close' to n,

as the nth root of Z^h is close to Z, less than Z+1 , in fact.

If we now use some information about square roots, the importance of there being a Z^h becomes significant.

Because of the equation between Z^h and X^h, Y^h, if we increase the left hand side by a power to make Z^h*m = Z^n, according to Pythagorus theorem, we would would have to increase both X, Y by more than than a factor of m in order for the relationship to still hold. Z is a larger number than X,Y , so increasing Z by a factor, means that X,Y will also increase but by the same factor that will make X,Y larger but not enough to be equal to a whole number.

If you increase the hypotenuse by a factor m, the 2 legs of the triangle have to increase by more than a factor m, in order that the relationship remain the same, ie. that whole numbers remain whole numbers.

So if Z^h = X^h + Y^h then Z^h*p = (X^n*m + Y^n*m) where p > 1, m > p, if Z,XY remain whole numbers. Because we chose to look at Z^n +1, the nth root is very close to Z. This also implies that nth roots of X^n*m and Y^n*m are very close to X,Y.

So for all cases of Z^n, n > 2, X^n + Y^n < Z^n < X^n + Y^n +1
It is impossible for Z^n to be a whole number, being between two consecutive whole numbers.

visual diagram of proof

This is the section of number line where x,y, z, x+y located relative to each other (order is important, not distance)

_______ X ________ Y ___ Z ___ X + Y ____

Here is section of number line where X^2 to (X+Y)^2, with Y^2, Z^2, (X^2 +Y^2), etc


___ X^2 ___Y^2__ (X^n + Y^n)^2/n __ (X^2 + Y^2)^2/2


Now Z^2 =(X^n + Y^n)^2/n and is less than (X^2 + Y^2)^2/2 =(X^2 + Y^2)^1

So Z^2 < (X^2 + Y^2)^1

There must exist a value, call it a, where 0 < a < 1 where

Z^2 = (X^2 + Y^2)^a

But if X,Y, and Z are whole numbers, then the left hand side of equation is a whole number, while the right hand side is an non-whole number.
This is true because a whole number raised to a non-whole number is an non-number number.

Thus this contradiction proves there is no X,Y,Z all being whole numbers that can meet this requirement.

another counter-example

Let's look at the familiar equation 3^2 + 4^2 = 5^2

What happens to 3^3 + 4^3 = 73 ? The cube root of 73 is 4.17 (approximately)

What happens to 3^4 + 4^4 = 291 ? The 4th root is 4.13 (approximately.

The progression of the higher roots is toward the number 4.
5, 4.27, 4.13 As the root gets higher in number, the root gets closer to 4. While the square root may be greater than y + 1, eventually the distance between z and y is less than 1. this means that there is a finite number of possible solutions for z, when x,y are fixed. So there are only a finite number of steps to consider for each z.

some counter examples

Here is one example that shows part of the process.

I claim x,y are 2 different numbers.
suppose x=y then X^n + X^n = Z^n

or 2X^n= Z^n

This implies that either x or Z must have a factor of 2^1/n as part of the number. One of them is irrational.

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