### Thursday, October 31, 1996

## Proof Correct

Thanks to the help of an open minded mathematician I've been able to verify that my proof of Fermat's Last Theorem is correct.

At this point, I no longer have a need to post the proof on the Net; although, I do hope to do so with the final draft of it at a later date.

It's been fun.

At this point, I no longer have a need to post the proof on the Net; although, I do hope to do so with the final draft of it at a later date.

It's been fun.

## To whomever is annoyed by my posts, Zen meditations

I admit up front that I'm an amateur trying to communicate with the professionals, so I have to do things differently.

I started on FLT for many reasons, not the least of which is is an excellent way to break down the ego: Nothing like a simple to formulate impossible (or nearly impossible) problem to prove to bring down high opinions.

I considered it a better koan than the usual stuff you get with Zen (like "what is the sound of one hand clapping").

Well, like many people, I came up with something that I thought answered it and excitedly went to mathematicians to tell somebody. I think you know the reception I got. Well, that just made me more determined and I was able to get greater access to the editors of AMS than many of you have ever had. And, I had nothing.

Well, I went back to my koan, and found that my escapades were great for breaking down my ego and were embarrassing besides. Luckily, some of you were kind enough to send me some nice words which I deeply appreciate.

Then, about a week ago, having learned much more about this problem, I was again confronted with a solution (which actually blows this as a koan, they aren't supposed to have solutions). So I made posts hoping that someone would help me like before.

I started on FLT for many reasons, not the least of which is is an excellent way to break down the ego: Nothing like a simple to formulate impossible (or nearly impossible) problem to prove to bring down high opinions.

I considered it a better koan than the usual stuff you get with Zen (like "what is the sound of one hand clapping").

Well, like many people, I came up with something that I thought answered it and excitedly went to mathematicians to tell somebody. I think you know the reception I got. Well, that just made me more determined and I was able to get greater access to the editors of AMS than many of you have ever had. And, I had nothing.

Well, I went back to my koan, and found that my escapades were great for breaking down my ego and were embarrassing besides. Luckily, some of you were kind enough to send me some nice words which I deeply appreciate.

Then, about a week ago, having learned much more about this problem, I was again confronted with a solution (which actually blows this as a koan, they aren't supposed to have solutions). So I made posts hoping that someone would help me like before.

### 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.

### Wednesday, October 23, 1996

## Stating the obvious. More on this FLT thing

I pretty much realize that it's probably well known and obvious that with FLT, neither x,y nor z can be divisible by n or there's a nice contradiction. It was fun proving it to myself anyway.

Here's something else that I haven't read but I guess is well established. I call it a strong case of Fermat's Little Theorem.

Given a-b divisible by n, a^n - b^n must be divisible by n^2

So Fermat's Last Theorem can be written as

Given x,y,z relative prime; n odd prime

no solutions exist unless (x+y)^n - (x^n + y^n) is divisible by n^2

Wonderfully obvious but new to me. Now I've noticed that x and y can be written in terms of n like x=an+f and y=bn+g. If (x+y)^n - (x^n + y^n) were divisible by n^2 this would require that f^n + g^n equal some h^n. Seems there's some infinite regression sort of thing.

Notice that for n=5 the above equals 5xy(x+y)(x^2 + xy + y^2) which would mean that x^2 + xy + y^2 must be divisible by n. It's easy enough to see that it can't be.

Here's something else that I haven't read but I guess is well established. I call it a strong case of Fermat's Little Theorem.

Given a-b divisible by n, a^n - b^n must be divisible by n^2

So Fermat's Last Theorem can be written as

Given x,y,z relative prime; n odd prime

no solutions exist unless (x+y)^n - (x^n + y^n) is divisible by n^2

Wonderfully obvious but new to me. Now I've noticed that x and y can be written in terms of n like x=an+f and y=bn+g. If (x+y)^n - (x^n + y^n) were divisible by n^2 this would require that f^n + g^n equal some h^n. Seems there's some infinite regression sort of thing.

Notice that for n=5 the above equals 5xy(x+y)(x^2 + xy + y^2) which would mean that x^2 + xy + y^2 must be divisible by n. It's easy enough to see that it can't be.

## Reg FLT, proof that x,y,z not divisible by n

As a hobby I've been playing around with FLT partly because it is considered a hopeless quest. That way, I knew I could probably play with it forever.

However, I just stumbled upon something that seems profound to me so I thought I'd toss it here so that someone give it a quick critique. This is not long.

Given all that usual stuff about x^n + y^n = z^n

let x=af, y=bg, z=ch where 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. And, (x+y-z)^n = n(z-x)(z-y)(x+y)Q where Q is all those other terms that are hard to write for the general case. For n=3 it is one. Also I have Q=q^n

Trying z divisible by n and using the above I get the following which must be true.

f^n + 2nfghq + g^n = n^{n-1}h^n subtracting f^n + nfgp + g^n = (f+g)^n

gives

nfg(p-2hq) = (f+g)^n - n^{n-1}h^n

p is always divisible by (f+g) because nfgp=(f+g)^n - (f^n + g^n)

since (f+g) must be divisible by n, because z is divisible by n, requires that h or q be divisible by n which is a contradiction.

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 or q be divisible by n because (g-h) is divisible by n which is a contradiction.

For n>3 with neither x,y or z divisible by n the above is just as conclusive. Therefore, I'm stuck wondering what could be wrong with the above only one of you will tell me.

Supposing that z is divisible by n, then z=nch and x+y=n^{n-1}h^n

However, I just stumbled upon something that seems profound to me so I thought I'd toss it here so that someone give it a quick critique. This is not long.

Given all that usual stuff about x^n + y^n = z^n

let x=af, y=bg, z=ch where 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. And, (x+y-z)^n = n(z-x)(z-y)(x+y)Q where Q is all those other terms that are hard to write for the general case. For n=3 it is one. Also I have Q=q^n

Trying z divisible by n and using the above I get the following which must be true.

f^n + 2nfghq + g^n = n^{n-1}h^n subtracting f^n + nfgp + g^n = (f+g)^n

gives

nfg(p-2hq) = (f+g)^n - n^{n-1}h^n

p is always divisible by (f+g) because nfgp=(f+g)^n - (f^n + g^n)

since (f+g) must be divisible by n, because z is divisible by n, requires that h or q be divisible by n which is a contradiction.

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 or q be divisible by n because (g-h) is divisible by n which is a contradiction.

For n>3 with neither x,y or z divisible by n the above is just as conclusive. Therefore, I'm stuck wondering what could be wrong with the above only one of you will tell me.

Supposing that z is divisible by n, then z=nch and x+y=n^{n-1}h^n