This is a proposed outline of a proof of Fermat's Last Theorem:
No diophantine equation (in integers) exists for x^n + y^n = z^n for n > 2.
Author: Gregory N. Mazur, www.tell-all.com
First Update: December, 1999
2nd Update: July, 2000 Please point out the flaws. Use either e-mail or the fast forms I supplied on the linked "feedback" page.
Let z and n be fixed integers with n > 2.
Let z^n = 2r^n such that r is a fixed "constant."
1) Define variables v and u within a fixed range such that:
2) z^n = ( r + v )^n + ( r - v )^n - u
3) The smallest absolute values in the range:
4) for v = 0, u = 0
5) The largest values in the range:
6) Where ( r + v )^n = 2r^n then:
the largest v = -r + nth root (2r^n) = D
7) Given D, then the largest u is the negative term, -u = -U, and
U = ( r + D)^n + ( r - D)^n - z^n where
8) (r - D)^n - U = 0 or
9) U = (r - D)^n
-U = - (2r - nth root(2r^n))^n
10) So the right handed possitive v "variable" varies from v = -r through v = +D.
11) The left handed u variable ranges from -u through -U
When v = -r, then -u = -(2r)^n + (2r^n)
12) Differentiating the equation with respect to u where z^n = (r + v)^n + (r - v)^n - u we have:
13) d(z^n)/du = 0 = n(r + v)^n-1(d(r + v)/du) + n(r - v)^n-1(d(r - v)/du) - du/du
14) Since r is a fixed constant, we have:
0 = n(r + v)^n-1(dv)/du)+ n(r - v)^n-1(d(-1*v)/du) - 1
15) 1 = dv/du * (n)((r + v)^n-1 - (r - v)^n-1)
16) dv/du = 1/n * ((r + v)^n-1 - (r - v)^n-1)
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17) Since z^n = (r + v)^n + (r - v)^n - u,
then let:
18) x = r + v;
19) y^n = (r - v)^n - u;
20) y = nth root ((r - v)^n - u)
21) y = ((r - v)^n - u)^1/n
22) the relationship of y to u and v is:
dy/du = (1/n)*((r-v)^n -u)^((1/n)-1)(d((r-v)^n -u)/du)
dy/dv = (1/n)*((r-v)^n -u)^((1/n)-1)(d((r-v)^n -u)/dv)
23) We *know* that r cannot be an integer when the integer z^n = 2r^n since r = nth root((z^n)/2)
24) When x must be an integer by definition, then v = the non-integer (x - r).
25) For any integers, z^n and x^n we *can* compute v:
26) The *FIRST* integer larger than r or the value (r+v) = x_1.
27) The value v is the sum of *TWO* values.
28) Let s = the size of the difference between the first integer larger than r and...
the fixed r such that (s < x_1 < D):
29) Since x_1 is the first larger integer, the 2nd component of v = 0:
v_0 = s + 0
30) For the next sequential integers greater than x_1, viz. x_2, x_3..., the v value has two known components:
31) The computed s remains unchanged for v_1, v_2, v_3...v_D.
32) The *integer* component of v_1 is the value q = (x_2 - x_1) = (r+v)/x = (x^n + y^n)/z^n
v_1 = s + q
v_2 = s + 2q
v_3 = s + 3q
.
.
.
v_D = (s + mq) = D where the largest v = D = -r + nth root (2r^n)
33) Now (r + s) must be an integer for q=0 by definition because...
x^n is an integer by definition.
34) But...(r - s) cannot be an integer because...
s < > q/2 where n >2.
For example, if r is short of an integer by "segment" s (s < 1/2)...then...
(r - s) is greater than an integer by (1 - s) which is more than 1/2
But... (r - s) must exceed an integer by the *same* amount that r was short of an integer for (r - s) to *BE* an integer. So (r - v) cannot be an integer because (r - s) cannot be... at least for an for n > 2.
There is a fundamental reason why s cannot be exactly 1/2 of ("q" = 1)...for n >2
The s "integer difference" has a different basis of representation than its companion r for n > 2.
r = nth root((2r^n)/2)...where x = r + s + mq.
s = (x - mq)- nth root((2r^n)/2)
35) MY FIRST premise: *IF* (r+d) = x must be an integer, then... (r-v) cannot be an integer.
For example, when r and s have the same basis of representation, then...
both must be multiples of the same basis which is (1/sqrt(2))... in the case of *integer* q = "1" when x, y, z and n *must be* integers as in a diophantine version of x^2 + y^2 = z^2.
Sqrt(2) can be defined as the EQUAL halves: 1/sqrt(2) + 1/sqrt(2) where 1/sqrt(2) divides all terms in: Sqrt(1^2 + 1^2) = Sqrt(Sqrt(2)^2)
Similarly:
nth root(r^n + r^n) = nth root(2r^n) where r = nth root(2r^n)/2) by definition.
As 1^2 + 1^2 = 2 redefines 1 and 2 as functions of the base number 1/Sqrt(2)...
Also r and z get redefined as *FUNCTIONS* of a NON-intger segment of z^n such as the nth root of the NON-integer (1/(2r^n)/2)."
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36) dv/du - 1/n*((r + v)^n-1 -(r - v)^n-1) as noted in step 16 above. Given x = (r + v) we have:
37) d(x - r)/du = 1/n*((r + v)^n-1 - (r - v)^n-1) where r is a *constant.* d(x - r)/du = dx/du + 0.
38) dx/du = dv/du = 1/n*((r + v)^n-1 - (r - v)^n-1)
39) u = (r - v)^n - y^n where...
40) y is a dependent variable for fixed integers z, n and the fixed non-integer, r in:
z^n - (r + v)^n = y^n
41) du/dv = n(r- v)^n-1(d(r-v)/dv - d(z^n - (r+v)^n)/dv
42) du/dv = n(r- v)^n-1(0 - 1) - (0 - n(r+v)^n-1(d(r+v)/dv)
43) du/dv = n(r-v)^n-1(0 - 1) - (0 - n(r+v)^n-1(0 + 1)
44) du/dv = -n(r-v)^n-1 + n(r+v)^n-1
45) du/dv = -n((r-v)^n-1 - (r+v)^n-1)
46) Every point P(v,u) will have a *solution* such that:
v = 1/n*((r+v)^n-1 - (r-v)^n-1)
u = -n((r-v)^n-1 - (r+v)^n-1)
47) y^n = (r-v)^n - u.
48) y^n = (r - 1/n*((r+v)^n-1 - (r-v)^n-1))^n + n((r-v)^n-1) - (r+v)^n-1)
49) Since the nth root of (r-v)^n *cannot* be an integer, then we conclude...
-u = n((r-v)^n-1 - (r+v)^n-1) *cannot* be an integer *because*...
the term -u is a "n" integer MULTIPLE of a NON-integer sum or n(a+b) for:
(a) NON-integer, (r-v)^n-1 and;
(b) Negative Integer -(r+v)^n-1.
50) MY SECOND premise: *IF* (r + v) = x must be an integer, *then* u cannot be an integer.
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51) Similarly, for integers x, z and n:
Recall y^n = (r-v)^n -u per 17,18,19. Also:
dy/du = (1/n)*((r-v)^n -u)^((1/n)-1)(d((r-v)^n -u)/du)
dy/dv = (1/n)*((r-v)^n -u)^((1/n)-1)(d((r-v)^n -u)/dv)
dx/du = dv/du = 1/n*((r+d)^n-1 -(r-d)^n-1) per 38.
52) Every point P_2(y,u) will have a *solution* such that:
y is an integer MULTIPLE of a non-integer;
53) Every point P_3(y,v) will have a *solution* such that:
y is an integer MULTIPLE of a non-integer;
54) Every point P_4(x,u) and P_5(v,u) will have a *solution* such that:
x is an *integer* by definition where (r+v) = x.
55) CONCLUSION:
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Since y must be an integer MULTIPLE of a non-integer when x, z and n are integers, then y cannot be an integer where x^n + y^n = z^n for n > 2.
Since y cannot be written as a unique product of primes, neither can a dependent variable x when a given y is a defined integer. Thus y and x cannot both be integers given integers z and n > 2 which is a PROOF of Fermat's Last Theorem.
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