Here is Fermat's famopus drawing that he could not put in the margin. See if you can work this out.

rewritten Fermats

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Fermat's Last Theorem – simple proof.

Let's assume X^n + Y^n = Z^n for X,Y,Z being rational numbers, with n > 2 and n= whole number.

For ease of demonstrating the argument, let n = 3. It will be easy to translate the argument for any n later.

The argument goes like this:

Because we assume X^n + Y^n = Z^n, we can make certain statements that are true about X, X^2, Y, Y^2, Z, Z^2. With these statements, we can solve Z in terms of a mix of numbers, including Y, and also X.

Most of the numbers are rational, due to assumptions or calculations. There is one number that it is impossible to determine if it is rational or irrational, specifically (m^3 + 1)^1/3 for the case n =3. Examples can be found where this number is either rational or irrational.

If it is irrational, then we are finished, as either Y or Z or both are irrational (m^3+1)^1/3. This is a contradiction with original assumption.

If it is rational, then we can rewrite the original equation X^3 + Y^3 = Z^3 as

(XX)^3 + (YY)^3 = (ZZ)^3, where XX, YY, and ZZ are all smaller than their counterparts X, Y, and Z by a factor of (m^3 + 1)^1/3, which we have chosen to be rational and which is larger than 1.

If we go through the same reasoning with this new equation, we again come to a new conclusion that (ZZ)/(YY) = (k^3 + 1)^1/3 , which while still larger than 1, the k in the equation is smaller than m.

If (k^3 + 1)^1/3 is irrational, then we are finished, as either ZZ or YY, or both are irrational.

But if (k^3 + 1) is rational, then this too can be factored out of the equation, and a new solution could be found, say j. (j^3 + 1)^1/3 is now rational or else there is a contradiction. Each time the factor is getting smaller, ie m> k > j.

It is not even necessary for there ever to be a stop to this process. There would be an infinite number of factors of Z, and you could never determine if there was even a repeating set of numbers, as the factors are continually getting smaller. But isn't this just a variation of the definition of an irrational number? If the factors of a number are infinite and different from each other, can there ever be a repeating set of numbers? There's always another factor waiting in the wings to disprove any collection of digits.

That is the argument. Here is the math.

If X^3 + Y^3 = Z^3, then X + Y = A*Z where 1<2^1/2 and X^2 +Y^2 = B*Z^2 where

1<2^1/2

reason ; X is chosen to be smaller of X,Y. Maximum value of X is less than that of Y, addition of both is less than 2 Y. If Z is greater than 2^1/12 Y, then Z^2 is greater than 2Y^2, and for every n>2 thereafter.

So, we have 3 equations now

1. X + Y = A*Z

2. X^2 + Y^2 = B*Z^2

3. X^3 + Y^3 = Z^3

Because X is smaller than Y (because we chose X to be smaller), let X = mY where 0<1.

Then our 3 equations become

1 m*Y + Y = A*Z

2 m^2*Y^2 = B*Z^2

3 m^3*Y^3 + Y^3= Z^3

3 becomes (m^3 + 1) Y^3 = Z^3

or further Z^3 / Y^3 = m^3 + 1

or Z/Y =(m^3 + 1)^1/3

If (m^3 + 1)^1/3 is irrational, then either Y, Z or both are irrational. So assume it is rational, and the argument above applies.

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.