### Wednesday, August 29, 2001

## Resistance to simple FLT proof boggles my mind.

If any of you have wanted me to explain how to prove Fermat's Last Theorem in a very simple way then this post is for you.

I show that x^p + y^p = z^p, requires that

(a1 z^2 + b1 xy)…(a2 z^2 + b2 xy) = 0(mod x^2 + y^2 + vz^2), where v is an integer, and a1...ap = v^p+1, b1...bp = -2, are just two of the p+1 defining equations for the 2p unknowns.

Then I notice something spectacularly simple.

I notice that x^2 + y^2 - z^2 MUST share prime counting number factors with x and y, if you assume x, y and z are nonzero integers.

And then I make use of the indeterminancy of the a's to consider a1 = sqrt(v+1) and a2 = sqrt(v+1), where v = -1 plus a multiple of some integer factor of x or y that is shared with x^2 + y^2 - z^2 (you know, like if x has a factor 11 that is also a factor z-y, then 11 would be the type of factor I'm talking about).

I set v to -1 plus a multiple of such a factor so that I can force

x^2 + y^2 + vz^2 to have that factor to any power I wish (like with the above 11^100293849, if I wanted too).

That then OBVIOUSLY forces

(a1 z^2 + b1 xy)…(a2 z^2 + b2 xy) to have that factor to the SAME power(again, with the above, 11^1000293849).

BUT WAIT!!!! x or y HAS that factor, and z CANNOT since it's pairwise coprime to x and y, so that leaves the a's!!!!!

BUT, I just said I've set only a1 and a2 to have that factor since they both equal sqrt(v+1), so all those factors must come from (a1 z^2 + b1 xy) and (a2 z^2 + b2 xy).

Well, did you notice something else? I ARBITRARILY considered a1 and a2 equal to sqrt(v+1). What about a1 = v+1? Now NONE of the other a's has that factor so everything is forced on the one (a1 z^2 + b1 xy).

Some of you should be falling on the floor in amazement that such a simple and extraordinary proof could have been undiscovered for so long, but most of you are now completely befuddled.

I am not going to bother looking at replies to my other posts, but I will look to see who replies to this one.

I will also post the link to my website, which is

http://communities.msn.com/ProofofFermatsLastTheorem&naventryid=104

in case you want to look at interesting details.

Luckily for me, despite the fact that most of you are so limited that you're still just reacting by reflex to believe I'm just a nut, this argument is so damn simple that soon enough someone will realize it is correct, and then some of you will get that sinking feeling in your gut as you contemplate how you reacted when it was presented to you :-).

Thank God for that! I didn't realize how closeminded modern mathematicians apparently are. You folks have to be forced by extreme measures to accept even simple stuff that falls just a little bit outside what you're used to.

That is, none of you are much fun at all! Sure hope the world never depends on your type discovering anything, or humanity will surely perish.

I show that x^p + y^p = z^p, requires that

(a1 z^2 + b1 xy)…(a2 z^2 + b2 xy) = 0(mod x^2 + y^2 + vz^2), where v is an integer, and a1...ap = v^p+1, b1...bp = -2, are just two of the p+1 defining equations for the 2p unknowns.

Then I notice something spectacularly simple.

I notice that x^2 + y^2 - z^2 MUST share prime counting number factors with x and y, if you assume x, y and z are nonzero integers.

And then I make use of the indeterminancy of the a's to consider a1 = sqrt(v+1) and a2 = sqrt(v+1), where v = -1 plus a multiple of some integer factor of x or y that is shared with x^2 + y^2 - z^2 (you know, like if x has a factor 11 that is also a factor z-y, then 11 would be the type of factor I'm talking about).

I set v to -1 plus a multiple of such a factor so that I can force

x^2 + y^2 + vz^2 to have that factor to any power I wish (like with the above 11^100293849, if I wanted too).

That then OBVIOUSLY forces

(a1 z^2 + b1 xy)…(a2 z^2 + b2 xy) to have that factor to the SAME power(again, with the above, 11^1000293849).

BUT WAIT!!!! x or y HAS that factor, and z CANNOT since it's pairwise coprime to x and y, so that leaves the a's!!!!!

BUT, I just said I've set only a1 and a2 to have that factor since they both equal sqrt(v+1), so all those factors must come from (a1 z^2 + b1 xy) and (a2 z^2 + b2 xy).

Well, did you notice something else? I ARBITRARILY considered a1 and a2 equal to sqrt(v+1). What about a1 = v+1? Now NONE of the other a's has that factor so everything is forced on the one (a1 z^2 + b1 xy).

Some of you should be falling on the floor in amazement that such a simple and extraordinary proof could have been undiscovered for so long, but most of you are now completely befuddled.

I am not going to bother looking at replies to my other posts, but I will look to see who replies to this one.

I will also post the link to my website, which is

http://communities.msn.com/ProofofFermatsLastTheorem&naventryid=104

in case you want to look at interesting details.

Luckily for me, despite the fact that most of you are so limited that you're still just reacting by reflex to believe I'm just a nut, this argument is so damn simple that soon enough someone will realize it is correct, and then some of you will get that sinking feeling in your gut as you contemplate how you reacted when it was presented to you :-).

Thank God for that! I didn't realize how closeminded modern mathematicians apparently are. You folks have to be forced by extreme measures to accept even simple stuff that falls just a little bit outside what you're used to.

That is, none of you are much fun at all! Sure hope the world never depends on your type discovering anything, or humanity will surely perish.

## FLT Proof, simple after all

I made an interesting mistake while discussing my proof of Fermat's Last Theorem on this newsgroup when I was overly restrictive with my own argument.

Based on my holding on to some overly restrictive assumptions, I cast doubt on one of my latest arguments, and even believed it was wrong myself, when it was not and the proof is correct, and quite simple.

Explaining the above won't take long, and the explanation will also show why there are no valid mathematical objections to the proof.

Ok, I take x^p + y^p = z^p, and show that

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2),

and what's new there are the a's, b's and that v. Well the v is an integer, and the a's and b's are somewhat defined by an expression that I won't show here, since I don't have to, to explain my argument.

What I will add about them is that

a1…ap = v^p + 1 and b1…b2 = -2.

Also, I said somewhat defined because it's the crux of this argument and the relates to the overly restrictive assumption I made. I kept forgetting that the a's and b's weren't completely determined yet.

The explanation is simple. There are p+1 coefficients in the main expression and as you can see 2p unknowns. That means that if you, say, determine all the b's, like b1 = -2, and have all the other b's equal 1, you have now determined the a's.

To highlight this fact, I've kept producing the example

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), with f, g and h integers, and even gave a more detailed example using

x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)

in addition to the (x+2)(x+2) that you're all familiar with.

In the case above I CHOOSE b1 = (1+sqrt(3)i)/2 and that forces everything else.

So, back to

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2).

For any factor of x^2 + y^2 - z^2, I can force v^p + 1 to have that same factor by picking v = -1 + mF, where m is some counting number, and F is that factor.

But here's the key point that was the basis of my proof until I forgot what I just told you above and bowed to the negative pressure, my a's aren't determined above, so I can have each one of them have a factor of F as a factor.

Yup, it's that simple because that forces the b's to have the same factor.

Not sure?

Consider 2x + m, where we'll deal with integers for simplicity. Is it possible for that expression to be divisible by 2 if m is not?

That's the lever that I'm using here, but some have essentially claimed that what's true for integers doesn't apply for other numbers.

And I said it was, and what made me really upset is that they then essentially claimed the issue wasn't resolvable, and claimed it was up to me to resolve it if it was.

Ok, I said that I could answer any mathematical objections in this post, and the simple resolution actually goes back to what these people have been claiming as support.

Think about

x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)

again, and consider that letting x = b1/a1 or b2/a2 gives you that familiar 2.

If a1, a2, b1, or b2 is in any way a fraction, however you define that past rational numbers, then then they can be put into a non-fractional form without changing the value of the expression from x^2 + 4x + 4.

Looking at the primary polynomial that fact should be trivial to you, but if you accept it, you have to accept that my proof of Fermat's Last Theorem is correct.

You see, the same kind of fulcrum that I have to make the proof is now being applied to you and your beliefs because of its simplicity.

Will you lie to yourself? Will you lie to the world by posting against me?

Oh yeah, in case you're saying, but what if x^2 + y^2 - z^2 only has 2 as a prime factor, since it has to be even, well then z^2 + y^2 - x^2 would have to be the same way, and so would z^2 + x^2 - y^2.

They can't all have only 2 as a prime factor, so they aren't all integers.

By the way, that fact is supported by the actual values for x,y and z that have been given, which I've talked about so much. Remember x=sqrt(6)-i, y = sqrt(6)+i, and z=sqrt(6), with x^3 + y^3 = z^3? You'll notice that factor of 2 popping up, but no other integer factors, as you will with any such example.

Based on my holding on to some overly restrictive assumptions, I cast doubt on one of my latest arguments, and even believed it was wrong myself, when it was not and the proof is correct, and quite simple.

Explaining the above won't take long, and the explanation will also show why there are no valid mathematical objections to the proof.

Ok, I take x^p + y^p = z^p, and show that

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2),

and what's new there are the a's, b's and that v. Well the v is an integer, and the a's and b's are somewhat defined by an expression that I won't show here, since I don't have to, to explain my argument.

What I will add about them is that

a1…ap = v^p + 1 and b1…b2 = -2.

Also, I said somewhat defined because it's the crux of this argument and the relates to the overly restrictive assumption I made. I kept forgetting that the a's and b's weren't completely determined yet.

The explanation is simple. There are p+1 coefficients in the main expression and as you can see 2p unknowns. That means that if you, say, determine all the b's, like b1 = -2, and have all the other b's equal 1, you have now determined the a's.

To highlight this fact, I've kept producing the example

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), with f, g and h integers, and even gave a more detailed example using

x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)

in addition to the (x+2)(x+2) that you're all familiar with.

In the case above I CHOOSE b1 = (1+sqrt(3)i)/2 and that forces everything else.

So, back to

(a1 z^2 + b1 xy)…(ap z^2 + bp xy) = 0(mod x^2 + y^2 + vz^2).

For any factor of x^2 + y^2 - z^2, I can force v^p + 1 to have that same factor by picking v = -1 + mF, where m is some counting number, and F is that factor.

But here's the key point that was the basis of my proof until I forgot what I just told you above and bowed to the negative pressure, my a's aren't determined above, so I can have each one of them have a factor of F as a factor.

Yup, it's that simple because that forces the b's to have the same factor.

Not sure?

Consider 2x + m, where we'll deal with integers for simplicity. Is it possible for that expression to be divisible by 2 if m is not?

That's the lever that I'm using here, but some have essentially claimed that what's true for integers doesn't apply for other numbers.

And I said it was, and what made me really upset is that they then essentially claimed the issue wasn't resolvable, and claimed it was up to me to resolve it if it was.

Ok, I said that I could answer any mathematical objections in this post, and the simple resolution actually goes back to what these people have been claiming as support.

Think about

x^2 + 4x + 4 = (a1 x + (1+sqrt(3)i)/2)(a2 x + (1-sqrt(3)i/2)

again, and consider that letting x = b1/a1 or b2/a2 gives you that familiar 2.

If a1, a2, b1, or b2 is in any way a fraction, however you define that past rational numbers, then then they can be put into a non-fractional form without changing the value of the expression from x^2 + 4x + 4.

Looking at the primary polynomial that fact should be trivial to you, but if you accept it, you have to accept that my proof of Fermat's Last Theorem is correct.

You see, the same kind of fulcrum that I have to make the proof is now being applied to you and your beliefs because of its simplicity.

Will you lie to yourself? Will you lie to the world by posting against me?

Oh yeah, in case you're saying, but what if x^2 + y^2 - z^2 only has 2 as a prime factor, since it has to be even, well then z^2 + y^2 - x^2 would have to be the same way, and so would z^2 + x^2 - y^2.

They can't all have only 2 as a prime factor, so they aren't all integers.

By the way, that fact is supported by the actual values for x,y and z that have been given, which I've talked about so much. Remember x=sqrt(6)-i, y = sqrt(6)+i, and z=sqrt(6), with x^3 + y^3 = z^3? You'll notice that factor of 2 popping up, but no other integer factors, as you will with any such example.

### Monday, August 13, 2001

## Game over.

Ok, so I couldn't solve a nearly four hundred year old math problem. LOL.

I certainly do hope that Wile's proof is correct, but I no longer really care.

I hope to never knowingly talk to a mathematician again (rude bunch).

And I feel an immense amount of relief.

I certainly do hope that Wile's proof is correct, but I no longer really care.

I hope to never knowingly talk to a mathematician again (rude bunch).

And I feel an immense amount of relief.

### Tuesday, August 07, 2001

## FLT Proof Discussion: Progess!!!

For some time posters have been getting away with using semantics as objections to my FLT proof. I finally got tired of it, so I'll learn to speak the lingo enough to shoot that down.

It shouldn't take more than a day or two, but the same rules apply.

Every time you folks force me to jump through hoops, you make it that much more certain that when the proof is finally acknowledged I will have that much more power to enact change.

Why do you think I make these kinds of posts?

They are markers for governments and university administrators.

Later, when one of you jumps up and hollers about some particular change that's going to be forced upon you, I'll have these as reference points to not only make the case that I was forced to go much further than would have been necessary if you all were ethical or competent, but also to explain why I'm so pissed.

At this rate, none of you math professors will be able to spit without getting written permission from your university president first.

It shouldn't take more than a day or two, but the same rules apply.

Every time you folks force me to jump through hoops, you make it that much more certain that when the proof is finally acknowledged I will have that much more power to enact change.

Why do you think I make these kinds of posts?

They are markers for governments and university administrators.

Later, when one of you jumps up and hollers about some particular change that's going to be forced upon you, I'll have these as reference points to not only make the case that I was forced to go much further than would have been necessary if you all were ethical or competent, but also to explain why I'm so pissed.

At this rate, none of you math professors will be able to spit without getting written permission from your university president first.

## FLT Proof Discussion: Progess!!!

For some time posters have been getting away with using semantics as objections to my FLT proof. I finally got tired of it, so I'll learn to speak the lingo enough to shoot that down.

It shouldn't take more than a day or two, but the same rules apply.

Every time you folks force me to jump through hoops, you make it that much more certain that when the proof is finally acknowledged I will have that much more power to enact change.

Why do you think I make these kinds of posts?

They are markers for governments and university administrators.

Later, when one of you jumps up and hollers about some particular change that's going to be forced upon you, I'll have these as reference points to not only make the case that I was forced to go much further than would have been necessary if you all were ethical or competent, but also to explain why I'm so pissed.

At this rate, none of you math professors will be able to spit without getting written permission from your university president first.

It shouldn't take more than a day or two, but the same rules apply.

Every time you folks force me to jump through hoops, you make it that much more certain that when the proof is finally acknowledged I will have that much more power to enact change.

Why do you think I make these kinds of posts?

They are markers for governments and university administrators.

Later, when one of you jumps up and hollers about some particular change that's going to be forced upon you, I'll have these as reference points to not only make the case that I was forced to go much further than would have been necessary if you all were ethical or competent, but also to explain why I'm so pissed.

At this rate, none of you math professors will be able to spit without getting written permission from your university president first.

### Wednesday, August 01, 2001

## Answering the critics

One thing that I find appalling is how easily some of the posters on this newsgroup distort the truth. Since they claim that I'm the one doing so, I will stick to the concrete from here on out in this post, so that you the reader can properly evaluate what's going on.

First off, let me return to the issue of rings.

My understanding is that some have claimed that you can't have a proof of anything without specifying the ring.

I have claimed that a particular number can be a factor without any other information except that the number is in a ring.

Rather than endlessly debate the issue, I presented a quadratic to explain my position:

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), with f, g and h integers.

(For instance, x^2 + 4x + 4 = (x+2)(x+2), where f=1, g=h=4, and a1=a2=1, and b1=b2=2.)

And them I repeatedly emphasized that DESPITE knowing that f,g and h are integers, you CANNOT specify a particular ring to cover all possibilities for a1, a2, b1 and b2.

However, my contention is that a1 and a2 are factors of f, DESPITE that, and similarly b1 and b2 are factors of h, despite the fact that you CANNOT specify a ring that covers all cases.

Some posters have apparently believed that you could simply say that a1, a2, b1 and b2 are algebraic numbers; however, I pointed out that a1 and a2 could be quaternions.

The issue has to do with use of factor in a general sense without knowing anything other than that you're in a ring.

Replies by posters to my posts seem to push the extreme position that it is not possible to have a proof without specifying a ring, so they would seem to indicate that a1 and a2 cannot be handled in a mathematical proof without specifying them to a particular ring (like saying a1=a2=1, as I did in my example before).

However, I'd like this position to be backed up by axioms, as it appears to me to be convention.

I think this point is a great place to start in resolving these discussions.

I would challenge those that reply to do so objectively, and without malice.

First off, let me return to the issue of rings.

My understanding is that some have claimed that you can't have a proof of anything without specifying the ring.

I have claimed that a particular number can be a factor without any other information except that the number is in a ring.

Rather than endlessly debate the issue, I presented a quadratic to explain my position:

fx^2 + gx + h = (a1 x + b1)(a2 x + b2), with f, g and h integers.

(For instance, x^2 + 4x + 4 = (x+2)(x+2), where f=1, g=h=4, and a1=a2=1, and b1=b2=2.)

And them I repeatedly emphasized that DESPITE knowing that f,g and h are integers, you CANNOT specify a particular ring to cover all possibilities for a1, a2, b1 and b2.

However, my contention is that a1 and a2 are factors of f, DESPITE that, and similarly b1 and b2 are factors of h, despite the fact that you CANNOT specify a ring that covers all cases.

Some posters have apparently believed that you could simply say that a1, a2, b1 and b2 are algebraic numbers; however, I pointed out that a1 and a2 could be quaternions.

The issue has to do with use of factor in a general sense without knowing anything other than that you're in a ring.

Replies by posters to my posts seem to push the extreme position that it is not possible to have a proof without specifying a ring, so they would seem to indicate that a1 and a2 cannot be handled in a mathematical proof without specifying them to a particular ring (like saying a1=a2=1, as I did in my example before).

However, I'd like this position to be backed up by axioms, as it appears to me to be convention.

I think this point is a great place to start in resolving these discussions.

I would challenge those that reply to do so objectively, and without malice.