Monday, September 28, 2009

Galois' Memoir: Lemma 2 (Galois Resolvent)

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Galois Resolvent Function

For any equation f(x) with distinct roots, the Galois Resolvent Function is a function g(x1, ..., xn) of the roots that no matter how the roots are permuted on the function, no two of the values are equal.

Definition 2: Galois Resolvent

The Galois Resolvent is a value of the Galois Resolvent Function where the roots of the equation f(x) are passed in as parameters.

Lemma 2: Galois Resolvent Function Exists

Given any equation f(x) with distinct roots a,b,c,... one can always form a function V of the roots such that no two of the values one obtains by permuting the roots in this function are equal.

For example, one can take:

V = Aa + Bb + Cc + ...

A, B, C, ... being suitably chosen whole numbers.


(1) Let the n distinct roots of f(x) be denoted a,b,c, ...

(2) Since these roots are distinct, the discriminant (a - b)2(a - c)2(b - c)2*... = D is not zero [For review of the discriminant, see here].

(3) What needs to be shown is that n integers A,B,C, ... can be chosen so that the n! numbers AS(a) + BS(b) + CS(c) + ... + are distinct where S ranges over all n! permutations of the roots a,b,c,...

[for details on why count(n permutations) = n!, see Corollary 1.1, here]

(4) Let P be the product of the squares of the differences of these n! numbers that is:

P = ∏ (S,T) [ A(S(a) - T(a)) + B(S(b) - T(b)) + ... ]2

where the product is all over n!(n! - 1)/2 pairs (unordered) of permutations S and T in which S≠ T.

Note: The purpose of this equation is to verify that all n! numbers are distinct. P is the product of all possible differences between any two permutations.

We know that there are n! possible permutations (see step #3 above).

We pick one of these permutations (1 out of n!) and call it S. Then, we pick a second one (1 out of n! - 1) and call it T. Since ordering doesn't matter and there are two ways to pick the same combination, we only need to deal with n!*(n!-1)/2 comparisons between S and T.

(5) To complete the proof, we only need to show that we can pick A,B,C, ... etc. such that P is nonzero.

If any of the permutations are not distinct, then the difference between S and T will be 0. If any of the differences are 0, then P = 0. So if P ≠ 0, it follows that we have found values for A,B,C,... such that all permutations are distinct.

(6) Let A, B, C, ..., be regarded at first as variables.

(7) Then P is a polynomial in variables A, B, C,... whose coefficients are polynomials in the roots a,b,c,...

(8) P is symmetric in the roots (this follows directly from the definition of P and the definition of symmetric polynomials, see Definition 1, here).

(9) Since P is symmetric, the coefficients are symmetric in the roots so using Waring's Method (see Theorem 4, here), P is a polynomial in A, B, C, ... with the coefficients symmetric in roots.

(10) So, we can determine the coefficients of P based on the elementary symmetric polynomials using the roots [see Theorem 1, here].

(11) We can therefore assume that the coefficients are known since we are assuming that the roots are known.

(12) Since P is a product of nonzero polynomials, it is nonzero. [see Theorem, here]

(13) Therefore once can assign integer values to A,B,C,... in such a way as to make P ≠ 0 [see Theorem, here].




Sugesh said...

I Sugesh krishna C.P. Thiruvanathapuram,Kerala ,INDIA have found a simple proof for Fermat's theorem. My e-mail id is

pdabous said...

8. P is symmetric w.r.t.variables A,B,C, . . . (the roots A', B', . . ) or w.r.t. a,b,c,. . . (the "roots" of f(x)?

12. Why is each factor of P a nonzero polynomial?

Alastair Bateman said...

The Marginal Proof of Fermat's Last Theorem

Step[i] Assume C^n = a^n + b^n exists for 4 natural / integer numbers C, a, b & n.
Step[ii] The difference of two squares is given by the well proven identity
x^2 - y^2 = (x + y)•(x - y) : Euclids Elements : Book 2 : proposition 5 & 6.
Step [iii] x & y can be any of the infinity of the natural numbers so let x = a^n and y = b^n therefore (a^n)^2 - (b^n)^2 = (a^n + b^n)•( a^n - b^n) hence a^n•a^n - b^n•b^n = C^n•( a^n - b^n).
By the Distributive Law of Multiplication: a^n•a^n - b^n•b^n = C^n•a^n - C^n•b^n
But C^n cannot be equal to both a^n & b^n at the same time
Hence the equation C^n = a^n + b^n has no reality when all four variables are natural / integer numbers as Pierre De Fermat stated all those years ago. Q.E.D.
As Fermat said "remarkable" because it is simple, absolute in what it proves, would almost fit in the margin of and is in perfect mathematical accord with where it was found in 'Arithmetica'.
A.R.Bateman :

Alastair Bateman said...

I forgot I had posted the comment. It is of course wrong because the equation (a^n)^2 - (b^n)^2 = C^n(a^n -b^n) is VALID!! C^n can seem to be equal to a^n & b^n at the same time because there is a profundity here that is not apparent at first sight. Since C^n = a^n + b^n then a^n(a^n + b^n) - b^n(a^n+b^n) = a^n + b^n ( what I call the 'escape knot equation' ) so we are back where we started and can go round the loop endlessly.( 'Infinite Descent'?) What this demonstrates is that all x^n ultimately collapse to a solution in the squares and as far as I can see it was the observation that the generic equation x^n = z^n2- y^2 is obtained when x^n is split because the algorithm is precisely the one used to split a square into two squares which is where Fermat made his observation in 'Arithmetica'. The result is always an all integer solution for x,y,z and the exponents but only two of the exponents are the same power which is always 2 so Fermat's statement was true, so yes he did have a proof contrary to what present day pundits tell us. That means as far as I am concerned that the quest for the modern proof is founded on the lie that Fermat did not have one. I have put a revised proof as a video on YouTube called 'Simpleton Proof of FLT' for the benefit of those who like me only understand high school maths.

Alastair Bateman said...

I am re-instating my original proof as valid with the following sentence inserted after REDUCTIO AD ABSURDUM. ' But a^n•a^n - b^n•b^n = (a^n + b^n)•a^n - (a^n + b^n)•b^n is valid since: (a^n•a^n + a^n•b^n) - (a^n•b^n + b^n•b^n) = a^n•a^n - b^n•b^n since the two 'strike through' terms cancel each other out. It has therefore been demonstrated that (a^n + b^n) only has an integer solution and can never equal C^n. '.

Thảo Trần said...

There is another explanation of a simple proof of Fermat’s last theorem as follows:
X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)

1. Let‘s divide (1) by (Z-X)^p, we shall get:
(X/(Z-X))^p +( Y/(Z-X))^p ?= (Z/(Z-X))^p (2)

2. That means we shall have:
X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =(X/(Z-X)), Y’ =(Y/(Z-X)), Z’ =(Z/(Z-X)) (3)

3. From (3), we shall have these equivalent forms (4) and (5):
Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)
Y’^p ?= p(-Z’) ^(p-1) + …+p(-Z’) +1 (5)

4. Similarly, let’s divide (1) by (Z-Y) ^p , we shall get:
(X/(Z-Y)) ^p +( Y/(Z-Y)) ^p ?= (Z/(Z-Y)) ^p (6)

That means we shall have these equivalent forms (7), (8) and (9):
X” ^p + Y” ^p ?= Z” ^p and Z” = Y”+1 , with X” =(X/(Z-Y)), Y” =(Y/(Z-Y)), Z” =(Z/(Z-Y)) (7)

From (7), we shall have:
X” ^p ?= pY”^(p-1) + …+pY” +1 (8)
X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X” ^p + Y” ^p ?= Z” ^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i) In (8) and (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1), that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY” ^(p-1)) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
Fermat’s last theorem is simply proved!

ii. With X^p + Y^p ?= Z^p , if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:
We should have : X^p + Y” ^p ?= Z” ^p , then X” ^p ?= 2Z” ^p or (X”/Z”) ^p ?= 2. The equal sign, in (X”/Z”) ^p ?= 2, is impossible.
Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”) ^p ?= 2. Is it interesting?

Alastair Bateman said...

To Thao Tran. 'Is it interesting?' Well for me personally it's a little, no a lot over my head as I am limited to OLD high school math's so I honestly couldn't say. Sorry.

Felipe Felipe said...

Hi Guys

I have found The Relations of Barlow when I read Ribenboim's Book. "Playing" with it, I found another sistem of equations, and then, I used modular congruences and I found what is summarized below (we can work just with natural numbers to analize the fermat's integer soluctions,because if one or 2 of the numbers are negatives, them we can change the place in the equation in order to have just positive numbers. If al of them are negative, we can multiply the equation by -1):

I didn't make an English version*, because I am having dificulties in the mahts expression (loses its format when I paste here), but if you follow the steps below, then you will achiev the same results :

* I published the entire demonstration in a Blog I created (in this blog is better to read the maths equations) and in other forums in Brazil and others countries :

From The Barlow Relations (See Fermat Last Theorem for Amateurs, pg 99 - but is very easy to show this) we have, in both Case 1 (n dont divide z) and Case 2 (n divide z) of FLT :

z-x = y0^n
z-y = x0^n

From this , we conclude thar

y-x = y0^n-x0^n = (y0-x0)(y0^(n-1)+....+x0^(n-1))

gdc(y-x, y0^(n-1)+....+x0^(n-1)) = y0^(n-1)+....+x0^(n-1)=p

y0^(n-1)+....+x0^(n-1) "=" 0 mod (p)
y-x "=" 0 mod (p)

Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n)

[(z-y)^(n-1)]^(1/n)+....+[(z-x)^(n-1)](1/n) "=" 0 mod(p)

Replacing y "=" x mod (p)

n[(z-x)^(n-1)](1/n) "=" 0 mod(p)

ny0^(n-1)"=" 0 mod(p)

p = 1 or p divides y0^(n-1) or p=n

As x and y are both > 0, then y0^(n-1)+....+x0^(n-1) is always > 1. So p can not be 1

If p divides y0, then it will divide x. But gdc(x,y,z)=1, so p can not divide y0 or x0

If p=n, then

y0^(n-1)+....+x0^(n-1)= n

But this sum has n terms, and this is possible just if y0=1 and x0 = 1 , because both are > 0. And if one of them is larger then 1, then the sum wil be larger than n.

If x0=1 and y0=1, then using the Relations of Barlow, we conclude that x=y, wich contradicts te conditions z>y>x and gdc (x,y)=1.

The same analysis could be used in the case 2 (even if n divedes x ou y).

So, there's no integers soluction for the equation x^n+y^n = z^n


Alastair Bateman said...

Hi guys,
After much deliberation I have at last seen the simplest and shortest proof of FLT which is so simple that it is laughable that it has taken us all so long to find it!
[1] Assume Z^n = Y^n + X^n exists.
[2] Multiply through by Z^n to give (Z^n)^2 = (Z.Y)^n + (Z.X)^n = (Z^2n)^1.
[3] Multiply through by Z^2n to give (Z^2n)^2 = (Z^3.Y)^n + (Z^3.X)^n.
[4] Multiply through by Z^2n to give (Z^2n)^3 = (Z^5.Y)^n + (Z^5.X)^n.
[5] We can continue like this ad infinitum to give every (Z^2n)^m. Every (Z^2n)^m forms a Pythagorean triple, not necessarily primitive so from step [2] onwards we have a one to one correspondence between the Fermat equation and the Pythagorean equation. So we can only conclude that the only possible value for n is 2 so that our assumption in [1] is an absurdity. Q.E.D.
nb: The simple procedure of splitting the square generates the Pythagorean triples from every (Z^2n)^m.
Although [2] seems very trivial it is a consequence of Euclids 'Elements' book 2 propositions 4 & 7 and which therefore constitute a proof for the 'Distributive Law of Multiplication'.
It is also a consequence of the 'Unique Composite Equation for the Triple of the Powers' i.e. X^n.Y^n.Z^n. which is equal to three composite equations.
All three methods generate three unique linked quadruples which are satisfied by the Pythagorean triples as inputs and since nobody else seems aware of them I have called them Fermat-Bateman quadruples.
81^2 = 9^3 + 18^3 at first sight seems to contradict [2] but of course 81^2 does not form a cubed integer and if it did it would prove Fermat's Last Theorem false.
Comments as to any flaws in my logic would be appreciated.