Wednesday, October 30, 1996
No real challenges yet to proof of Fermat's Last Theorem
So far, no one has come up with a real error in this latest rough draft of my proof of Fermat's Last Theorem. I did here of one area of possible confusion:
I show that x+y-z = nfghq
Note, that expression doesn't say which of x,y,z or Q is divisible by n.
For example, if x is divisible by n, then x=naf, (z-y)=n^{n-1}f^n
(x+y-z)^n = n(n^{n-1}f^n)(g^n)(h^n)(q^n)
(x+y-z) = nfghq
The method for the proof is correct. Of that I am certain.
I received email with cogent criticisms of my previous posts concerning
the difficulties in reading it. This post is a first attempt to put it
all in more readable format. I've included a definitions section to
explain more of what I simply stated before.
Despite the fact that this is still a rough draft it should be possible
now to follow it to its conclusion, and see that the method proves
Fermat's Last Theorem.
Further, constructive criticism will be appreciated.
James Harris
------------------------------------------------------------------
Introduction.
Fermat's Last Theorem has long been a magnet to the amateur and
professional mathematician alike because of its seeming simplicity; yet,
extraordinary difficulty. Although there is a proof by Andrew Wiles, I
think it is understandable that the problem still would incite
curiosity. I would also assume that a simpler solution would also be of
interest.
1. Statement of the Problem: Fermat's Last Theorem
Given x,y,z, relatively prime, n odd prime
no solution exists for the equation x^n + y^n = z^n
2. Definitions.
f=(z-y), g=(z-x), h=(x+y);
Now using the substitution for y with n=3 gives
x^3 = 3fy^2 + 3f^2 y + f^3 from which it is obvious by inspection
that x^3 is divisible by
f only once, except for possibly one other factor of 3. In general,
that also requires that those factors of f be raised to the nth power
except for the n factor. So now, using a,b,c for the other factors
x=af or naf, y=bg or nbg, z=ch or nch
x+y=h^n or n^{n-1}h^n, z-x=g^n or
n^{n-1}g^n, and z-y=f^n or n^{n-1}f^n.
Now in general, (x+y-z)^n = n(z-x)(z-y)(x+y)Q + x^n + y^n - z^n
where Q represents all those other terms that are hard to write out for
the general case. For
n=3 it is one. And for n=5
Q = z^2 - (x+y)z + x^2 + xy + y^2
I also use q where Q = q^n or n^{n-1}q^n
Finally, I use a term p, where nfgp = (f+g)^n - (f^n + g^n)
From which it is obvious that p is divisible by (f+g)
-------------------------------------------------------------------------
Now by Fermat's Little Theorem, since x^n + y^n = z^n,
x^n - (z-y) = 0(mod n), therefore x^n - f^n = 0(mod n), x - f = 0(mod n)
and also y - g = 0(mod n) and z - h = 0(mod n)
And therefore that (a - 1) = 0(mod n) with similar relations for b and
c
-------------------------------------------------------------------------
2. General Proof
Now from before remembering that x^n + y^n = z^n
(x+y-z)^n = n(z-x)(z-y)(x+y)Q which gives
(x+y-z) = nfghq
To use the above it is easier to consider individual cases where one of
x,y or z is divisible by n or none of them are. Note that if z is
divisible by n, then (x+y) and therefore (f+g) must be divisible by n,
as explained in the definitions section.
a. Case where z is divisible by n
x + y-z = x-f^n ; y + x-z = y - g^n; x+ y -z = n^{n-1}h^n-z = nfghq
Adding them gives
x+ y -z - f^n - g^n + n^{n-1}h^n = 3nfghq
Which is just
-f^n - g^n + n^{n-1}h^n = 2nfghq
From which I get
f^n + 2nfghq + g^n = n^{n-1}h^n and subtracting f^n + nfgp +
g^n=(f+g^)n
nfg(p-2hq) = (f+g)^n - n^{n-1}h^n
neither f or g or q can be divisible by n, because that would mean that
x or y would have to share a factor of n with z which violates the
condition of relative primeness. Therefore, the above requires that h
be further divisible by n.
-----------------------------------------------------------------------
The same comes up with x or y divisible by n since you get
g^n + 2nfghq - h^n = n^{n-1}f^n subtracting
g^n + nghp - h^n = (g-h)^n
gives
ngh(p-2fq) = (g-h)^n - n^{n-1}f^n
which requires that f,g,h or q be divisible by n because (g-h) is
divisible by n which again requires that f be further divisible by n for
the reasons stated before.
3. Proof for Case x,y,z not divisible by n
-------------------------------------------------------------------------
lemma(1): Extension of Fermat's Little Theorem:
Given a-b divisible by n, a^n - b^n must be divisible by n^2
Proof: a-b divisible by n implies that a equals some jn+r and b
equals some kn + r
Then, a^n - b^n = (jn)^n +...+n(jn)r^{n-1} + r^n -
[ (kn)^n +...+n(kn)r^{n-1} + r^n]
by inspection r^n can be subtracted off and the other terms are
multipled by n^2
-------------------------------------------------------------------------
So by lemma(1) if neither x,y nor z are divisible by n and q is
then
(x+y)^n - (x^n + y^n) must be divisible by n^2.
But since f has the same modulus with respect to n as x and g has the
same as y, it also requires that (f+g)^n - (f^n +g^n) be divisible by
n^2.
and from before
(f+g)^n - (f^n + g^n) = nfgp and p is seen to be required to be
divisible by n
But like before I can write
f^n + 2nfghq + g^n = h^n and again subtract f^n + nfgp + g^n =(f+g)^n
which gives
nfg(p-2hq) = (f+g)^n - h^n
and since both sides are divisible by n^2, q is then forced to be
further divisible by n.
4. Case for (x+y-z) divisible by n^2 or higher powers of n
Now then, these cases force me to rewrite my expression for (x+y-z).
Considering the situation with divisibility by n^2, I have
(x+y-z)=(n^2)fghq
-------------------------------------------------------------------------
But still for my other relations at least two aren't divisible by n.
Considering the possibility x,y not divisible by n
x+y-z = af - f^n = (n^2)fghq
This requires that af - f^n be divisible by n^2. But if I rewrite
using moduli
f=kn+r a=mn+1 giving (kn+r)(mn+1) - (kn+r)^n
Expanding (kn+r)^n all terms are divisible by n^2 except the last r^n
term
Now (kn+r)(mn+1) = km(n^2) + (rm + k)n + r , r
which requires that (rm+k) be divisible by n and that r^n - r be
divisible by n^2
-------------------------------------------------------------------------
But using s=n-r gives n^n -...+n(n)s^{n-1} - s^n - (n - s)
Which it can be seen requires that s^n - s + n must be divisible by n^2
-------------------------------------------------------------------------
So then by observation s^n - s can't then be divisible by n^2
But kn + r can be written as (k+1)n + (r - n) = (k+1)n - s
Expanding [(k+1)n-s]^n all terms are divisible by n^2 except the last
s^n term
and like before [(k+1)n-s](mn+1) = (k+1)m(n^2) - [sm - (k+1)]n - s , s
and as before [sm - (k+1)] must be divisible by n and s^n - s must be
divisible by n^2.
-------------------------------------------------------------------------
So I try, f = k(n^2 )+ r^n, a = m(n^2) + 1
Now if y is also not divisible by n and I have z divisible by n, then I
can also write
bg - g^n = (n^2)fghq
And the same argument applies so that g and b must be in the same form
as f and a.
For instance, g = j(n^2) - r^n, from which f+g = (j+k)(n^2)
But then (f+g) and therefore p would have to be divisible by (n^2) and
I would have
f^n + 2(n^2)fghq + g^n = n^{2n-1}h^n and subtracting
f^n + nfgp + g^n = (f+g)^n giving
nfg(p-2nhq) = (f+g)^n - n^{2n-1}h^n
which would require both sides to be divisible by n^{2n-1}.
Since p is divisible by n^2 this still requires that h be further
divisible by n, and a similar argument applies for x or y divisible by
n^2 like before.
Also, if neither x,y nor z is divisible by n, then f,g and h would all
have to be of the above format and a similar expression would require
that q be further divisible by n.
-------------------------------------------------------------------------
In general, whatever power t of n I use, p will be divisible by n to
that factor and the term on the other side will always have to be
further divisible by n.
nfg[p-2(n^{t-1})hq] = (f+g)^n - n^{tn-1}h^n
------------------------------------------------------------------------
Then it can be seen that n has to always continuously be raised to a
higher power and since there are an infinite number of n's there is no
integer solution, and Fermat's Last Theorem is proven.
I show that x+y-z = nfghq
Note, that expression doesn't say which of x,y,z or Q is divisible by n.
For example, if x is divisible by n, then x=naf, (z-y)=n^{n-1}f^n
(x+y-z)^n = n(n^{n-1}f^n)(g^n)(h^n)(q^n)
(x+y-z) = nfghq
The method for the proof is correct. Of that I am certain.
I received email with cogent criticisms of my previous posts concerning
the difficulties in reading it. This post is a first attempt to put it
all in more readable format. I've included a definitions section to
explain more of what I simply stated before.
Despite the fact that this is still a rough draft it should be possible
now to follow it to its conclusion, and see that the method proves
Fermat's Last Theorem.
Further, constructive criticism will be appreciated.
James Harris
------------------------------------------------------------------
Introduction.
Fermat's Last Theorem has long been a magnet to the amateur and
professional mathematician alike because of its seeming simplicity; yet,
extraordinary difficulty. Although there is a proof by Andrew Wiles, I
think it is understandable that the problem still would incite
curiosity. I would also assume that a simpler solution would also be of
interest.
1. Statement of the Problem: Fermat's Last Theorem
Given x,y,z, relatively prime, n odd prime
no solution exists for the equation x^n + y^n = z^n
2. Definitions.
f=(z-y), g=(z-x), h=(x+y);
Now using the substitution for y with n=3 gives
x^3 = 3fy^2 + 3f^2 y + f^3 from which it is obvious by inspection
that x^3 is divisible by
f only once, except for possibly one other factor of 3. In general,
that also requires that those factors of f be raised to the nth power
except for the n factor. So now, using a,b,c for the other factors
x=af or naf, y=bg or nbg, z=ch or nch
x+y=h^n or n^{n-1}h^n, z-x=g^n or
n^{n-1}g^n, and z-y=f^n or n^{n-1}f^n.
Now in general, (x+y-z)^n = n(z-x)(z-y)(x+y)Q + x^n + y^n - z^n
where Q represents all those other terms that are hard to write out for
the general case. For
n=3 it is one. And for n=5
Q = z^2 - (x+y)z + x^2 + xy + y^2
I also use q where Q = q^n or n^{n-1}q^n
Finally, I use a term p, where nfgp = (f+g)^n - (f^n + g^n)
From which it is obvious that p is divisible by (f+g)
-------------------------------------------------------------------------
Now by Fermat's Little Theorem, since x^n + y^n = z^n,
x^n - (z-y) = 0(mod n), therefore x^n - f^n = 0(mod n), x - f = 0(mod n)
and also y - g = 0(mod n) and z - h = 0(mod n)
And therefore that (a - 1) = 0(mod n) with similar relations for b and
c
-------------------------------------------------------------------------
2. General Proof
Now from before remembering that x^n + y^n = z^n
(x+y-z)^n = n(z-x)(z-y)(x+y)Q which gives
(x+y-z) = nfghq
To use the above it is easier to consider individual cases where one of
x,y or z is divisible by n or none of them are. Note that if z is
divisible by n, then (x+y) and therefore (f+g) must be divisible by n,
as explained in the definitions section.
a. Case where z is divisible by n
x + y-z = x-f^n ; y + x-z = y - g^n; x+ y -z = n^{n-1}h^n-z = nfghq
Adding them gives
x+ y -z - f^n - g^n + n^{n-1}h^n = 3nfghq
Which is just
-f^n - g^n + n^{n-1}h^n = 2nfghq
From which I get
f^n + 2nfghq + g^n = n^{n-1}h^n and subtracting f^n + nfgp +
g^n=(f+g^)n
nfg(p-2hq) = (f+g)^n - n^{n-1}h^n
neither f or g or q can be divisible by n, because that would mean that
x or y would have to share a factor of n with z which violates the
condition of relative primeness. Therefore, the above requires that h
be further divisible by n.
-----------------------------------------------------------------------
The same comes up with x or y divisible by n since you get
g^n + 2nfghq - h^n = n^{n-1}f^n subtracting
g^n + nghp - h^n = (g-h)^n
gives
ngh(p-2fq) = (g-h)^n - n^{n-1}f^n
which requires that f,g,h or q be divisible by n because (g-h) is
divisible by n which again requires that f be further divisible by n for
the reasons stated before.
3. Proof for Case x,y,z not divisible by n
-------------------------------------------------------------------------
lemma(1): Extension of Fermat's Little Theorem:
Given a-b divisible by n, a^n - b^n must be divisible by n^2
Proof: a-b divisible by n implies that a equals some jn+r and b
equals some kn + r
Then, a^n - b^n = (jn)^n +...+n(jn)r^{n-1} + r^n -
[ (kn)^n +...+n(kn)r^{n-1} + r^n]
by inspection r^n can be subtracted off and the other terms are
multipled by n^2
-------------------------------------------------------------------------
So by lemma(1) if neither x,y nor z are divisible by n and q is
then
(x+y)^n - (x^n + y^n) must be divisible by n^2.
But since f has the same modulus with respect to n as x and g has the
same as y, it also requires that (f+g)^n - (f^n +g^n) be divisible by
n^2.
and from before
(f+g)^n - (f^n + g^n) = nfgp and p is seen to be required to be
divisible by n
But like before I can write
f^n + 2nfghq + g^n = h^n and again subtract f^n + nfgp + g^n =(f+g)^n
which gives
nfg(p-2hq) = (f+g)^n - h^n
and since both sides are divisible by n^2, q is then forced to be
further divisible by n.
4. Case for (x+y-z) divisible by n^2 or higher powers of n
Now then, these cases force me to rewrite my expression for (x+y-z).
Considering the situation with divisibility by n^2, I have
(x+y-z)=(n^2)fghq
-------------------------------------------------------------------------
But still for my other relations at least two aren't divisible by n.
Considering the possibility x,y not divisible by n
x+y-z = af - f^n = (n^2)fghq
This requires that af - f^n be divisible by n^2. But if I rewrite
using moduli
f=kn+r a=mn+1 giving (kn+r)(mn+1) - (kn+r)^n
Expanding (kn+r)^n all terms are divisible by n^2 except the last r^n
term
Now (kn+r)(mn+1) = km(n^2) + (rm + k)n + r , r
which requires that (rm+k) be divisible by n and that r^n - r be
divisible by n^2
-------------------------------------------------------------------------
But using s=n-r gives n^n -...+n(n)s^{n-1} - s^n - (n - s)
Which it can be seen requires that s^n - s + n must be divisible by n^2
-------------------------------------------------------------------------
So then by observation s^n - s can't then be divisible by n^2
But kn + r can be written as (k+1)n + (r - n) = (k+1)n - s
Expanding [(k+1)n-s]^n all terms are divisible by n^2 except the last
s^n term
and like before [(k+1)n-s](mn+1) = (k+1)m(n^2) - [sm - (k+1)]n - s , s
and as before [sm - (k+1)] must be divisible by n and s^n - s must be
divisible by n^2.
-------------------------------------------------------------------------
So I try, f = k(n^2 )+ r^n, a = m(n^2) + 1
Now if y is also not divisible by n and I have z divisible by n, then I
can also write
bg - g^n = (n^2)fghq
And the same argument applies so that g and b must be in the same form
as f and a.
For instance, g = j(n^2) - r^n, from which f+g = (j+k)(n^2)
But then (f+g) and therefore p would have to be divisible by (n^2) and
I would have
f^n + 2(n^2)fghq + g^n = n^{2n-1}h^n and subtracting
f^n + nfgp + g^n = (f+g)^n giving
nfg(p-2nhq) = (f+g)^n - n^{2n-1}h^n
which would require both sides to be divisible by n^{2n-1}.
Since p is divisible by n^2 this still requires that h be further
divisible by n, and a similar argument applies for x or y divisible by
n^2 like before.
Also, if neither x,y nor z is divisible by n, then f,g and h would all
have to be of the above format and a similar expression would require
that q be further divisible by n.
-------------------------------------------------------------------------
In general, whatever power t of n I use, p will be divisible by n to
that factor and the term on the other side will always have to be
further divisible by n.
nfg[p-2(n^{t-1})hq] = (f+g)^n - n^{tn-1}h^n
------------------------------------------------------------------------
Then it can be seen that n has to always continuously be raised to a
higher power and since there are an infinite number of n's there is no
integer solution, and Fermat's Last Theorem is proven.
# posted by James Harris @ 09:10 
