An Elementary Proof Of Both The Beal Conjecture And Fermat's Last Theorem.

[maze]

It's hard to give a good analysis of the proof, because, as written, there are a lot of holes in it. Now, we have pointed out some of those and to your credit you have made some decent arguments as to why they might not really be a problem (why they are "aviodable"). But then each patch up brings new issues, etc.

If you wish for us to give a good critique of the proof, I would recommend you rewrite it, using the following format:
Let z > 2 and assume there is a solution.

...proof...

contradiction. (eg: you have equations that must be true but also can't be true)

Therefore there is no solution for z > 2. QED

Note that there is no need to even bother considering z=1 or z=2 since they can be shown true with simple examples 1+2=3 and 3²+4²=5².

 

[wsalem]

In fact, to help me remain consistent with those ideals, I am "keeping score".
In other words, I am keeping a list of statements in this thread
that were either wrong or belligerent.

Keeping score! Huh. Just because I misread your proof and replied with incorrect analysis doesn't make your proof more likely true! Heck, suppose I was a complete idiot, that wouldn't make your proof correct either.

Also, you're misquoting me, in an unethical way to be sure!

As you can see, I have endured and absorbed many unprovoked insults
without even once "retaliating", and made only one "honest mistake".
.
.
Belligerent Statements Made By Others:
You should know that you're in deep trouble!...__________________(8)

Well, I actually said, referring to the problem of division by zero

if any step after (4) proved or assumed that T=c or T=1 then you should know that you're in deep trouble!

Maybe you can get around the division and not get into trouble, but how was that hostile, or belligerent. Oh, unless you got to keep score!

 

[jambaugh]

...
The "indeterminate form" that you [maze] mention appears in the context: 1^\frac{0}{0}=1

and is therefore not an "issue" because clearly, \frac{0}{0}\neq\infty.
...

If you look at your indeterminate form it does indeed go to both plus and minus infinity depending on whether T approaches c from above or below.

As maze demonstrated naively utilizing indeterminate forms can lead to more obvious contradictions such as 1=0 by themselves and thus must be most carefully utilized especially in a RAA proof. If you wish to use them I suggest you start with only well defined expressions such as your starting point but with T>c or T<c and then take the limit T-->c a la undergraduate calculus.

As soon as you write down and real valued expression which is either not defined at all or not defined for some (important) values of the variables then your "proof" will be rejected out of hand.

You may be on to something or you may be blinded by your excitement so take the critiques to heart and as has been mentioned stick to a very rigorous format. Introduce assumptions explicitly and do not reassign values to variables and especially constants without an explicit enumeration of cases.

Finally do not assume that just because others find fault with your proof do not assume they did not understand it. Start by assuming they understood all to well an error you made. Remember you have the burden of proof for your "proof".

 

[Don Blazys]

To: wsalem,

Qoting wsalem:

Keeping score! Huh. Just because I misread your proof and replied with incorrect analysis doesn't make your proof more likely true! Heck, suppose I was a complete idiot, that wouldn't make your proof correct either.

Also, you're misquoting me, in an unethical way to be sure![\I]

The entire sentence:

Quoting wsalem:

And no matter what you do, if any step after (4) proved or assumed that T=c or T=1 then you should know that you're in deep trouble!

was perceived as hostile because of the "go for the throat and finish him off"
attitude it conveys. Also, the phrase "And no matter what you do..." imparts a "finality"
that may even cause some impressionable young readers to abandon
any initial curiosity that they may have had for this truly fascinating enigma.

I felt that was unfair.

By the way, I didn't "misquote you in an unethical way"!
In order to avoid making that post unnecessarily long,
I "copied and pasted" only "short snippets",
with the understanding that anyone can simply "scroll back"
to the original post containing the entire statement.

Anyway, I edited your comment off my list.
Given my age, and the state of my health,
my chances of success are slim indeed,
so I need friends, not enemys.

Quoting wsalem:

Heck, suppose I was a complete idiot,
that wouldn't make your proof correct either.

Of course not. However, the same rule would also apply to Einstein
for even he was prone to making blunders!
Not always, but as a general "rule of thumb",
I keep my replies to "complete idiots" as short as possible,
or I simply don't reply at all.
Please note the length of the replies that I gave you.

Don.

 

[Don Blazys]

To: maze,

Quoting maze:

It's hard to give a good analysis of the proof, because, as written, there are a lot of holes in it. Now, we have pointed out some of those and to your credit you have made some decent arguments as to why they might not really be a problem (why they are "aviodable"). But then each patch up brings new issues, etc.

If you wish for us to give a good critique of the proof, I would recommend you rewrite it, using the following format:
Let z > 2 and assume there is a solution.

...proof...

contradiction. (eg: you have equations that must be true but also can't be true)

Therefore there is no solution for z > 2. QED

Note that there is no need to even bother considering z=1 or z=2 since they can be shown true with simple examples 1+2=3 and 32+42=52.

Well, I wouldn't go so far as to say that it's got "a lot of holes".
Basically, we encountered two "issues".
The "indeterminate forms" was one "issue",
and establishing that the values of x, y, z in the unfactored case
must be different from the values of x, y, z in the factored case
in order to maintain terms that are positive integers was the other "issue".

Both "issues" have now been demonstrated as being "non-lethal" to the proof.

There are no other issues!

Can I at least get a "tentative" agreement that this result is at least "interesting"?

To quote "Dr. Evil", can somebody "throw me a bone?"

You see, from my point of view, a "formal proof" of
The Beal Conjecture And Fermat's Last Theorem
isn't even necessary,
because the moment that we factor the equation:

c^z-b^y=a^x,

we get:

\left(c^\frac{z}{2}+b^\frac{y}{2}\right)\left(c^\frac{z}{2}-b^\frac{y}{2}\right)=a^x,

and since the above two equations can also be viewed as:

T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x

and:

\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(\left(\frac{T}{T}\right)c^\frac{z}{2}-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x,

where a, b, c are co-prime,
it is obvious that T=c must be allowable,
and that this, in turn, requires that we first "cancel out"
the expressions involving logarithms at z=1 and z=2
so that the logarithms no longer exist.

Oops, I proved it again!

You see, the properties of logarithms themselves require that
both the Beal Conjecture and Fermat's Last Theorem be true!

Thus, no lawer like "step by step argument" is required,
because the properties of logarithms automatically and inherently
don't allow x, y, z all greater than 2.

It's just a "straightforward result"! An unavoidable consequence of logarithms!
I keep looking at it, and I keep asking myself.. "what's there to argue?"
It is what it is!

The moment some high school kid factors c^z-b^y=a^x,
substitutes a couple of "Blazys terms" and asks:
"How can I get rid of those logarithms so that I can let T=c?"
he or she solves the worlds most famous math problem!

His or her teacher might insist that T=c be prohibited,
but then that kid would be right, and the teacher would be wrong!
Every teacher that I showed this to initially disagreed with me,
but now agrees...so there is hope.

It may take a while for this to "sink in" to the hearts and minds of the "math community",
but it is, in fact, a real "mindblower" when you finally realize, as I did,
that the solution the worlds most famous math problem is nothing more than
a straightforward and hitherto unsuspected and undiscovered property of logarithms!

This is only one reason that I believe, to the very core of my being,
that my "Blazys terms" deserve further investigation.

In fact, it's a tragedy that they are so poorly understood.

Come on, when all is said and done, isn't it wonderfull that in the end,
it turns out that both the Beal Conjecture and Fermat's Last Theorem
can be demonstrated as being true using only "their own terms"
rather than some very, very, very distantly related concepts and constructs
involving "modular forms", "eliptical curves" and even "imaginary numbers"?

Shouldn't such good news be welcomed rather than "swept under a rug"?

Don.

 

[Don Blazys]

To: jambaugh,

Thanks for your post.

I'm out of time now, but I will answer it the very next chance I get.

By the way, that's a pretty disturbing "icon" you got there!

Don.

 

[jambaugh]

To: maze,
...
Thus, no lawer like "step by step argument" is required,
...

But that is exactly what a mathematical proof is.

...
I keep looking at it, and I keep asking myself.. "what's there to argue?"
It is what it is!

The moment some high school kid factors c^z-b^y=a^x,
substitutes a couple of "Blazys terms" and asks:
"How can I get rid of those logarithms so that I can let T=c?"
he or she solves the worlds most famous math problem!

This is the crux of the matter. You are "getting rid of terms" by selecting one set of values for certain parameters in a particular expression.
But this does not prove that the original expression can't occur with other sets of values. Let me give you an analogy. You can't drive to Hawaii but that doesn't mean Hawaii doesn't exist. Negative proofs are notoriously difficult.

It may take a while for this to "sink in" to the hearts and minds of the "math community"
...
In fact, it's a tragedy that they are so poorly understood.
...
Shouldn't such good news be welcomed rather than "swept under a rug"?

Mathematical truth isn't going anywhere. If your method has merit it will win out. But from your above quotes you have gone beyond questioning whether you are correct to questioning why others don't sing your praises. This is premature. Again you must be wary of a.) being blinded by your own enthusiasm and more importantly
b.) having so much emotionally invested in your belief that you are unable to face the possibility that you are completely mistaken.

Let me tell you a story... Back before the internet was all pervasive I was active on the compuserve physics discusssion group. There was a fellow in there who was sure that he had "discovered" a means of storing practically unlimited energy via capacitors. His mistake was in confusing charge with energy and did not appreciate that his method increased charge by sacrificing voltage (the product of which gives energy). I took him "by the hand" so to speak step by step through the calculations and he momentarily had a grasp of the issue that the energy per volume will be constant and not unbounded. But he suddenly stopped listening and fixated on the idea "I know I'm right!" He had spent years on this idea and was so emotionally invested he couldn't accept the possibility that all that investment was wasted. I occasionally see his posts around the web where he rants about "conspiracies" to keep his "discovery" suppressed.

So beware your own attitude. Remember changing your attitude will not change the truth or falsehood of what you claim only you ability to see this clearly. Take some time to simply conceptualize the possibility that you are grossly and embarrassingly wrong. Don't think about the problem itself just your attitude about the problem. Remind yourself that it is not going anywhere, if it is true it is true if it is wrong it is wrong. Get to the level of humility where you can accept the idea of being grossly wrong and then approach your proof as if you are trying to find and understand your error. When you have this frame of mind you can then look at the argument, as they say "without passion or prejudice".

Remember those "lawyer like" step by step arguments are insisted upon for a reason, and that reason is the same in mathematics as it is in the legal system... to get at facts independent of personal bias and prejudice. You are sure you are "right" or at least "onto something" and that is a bias you must fight when trying to refine and present your proof.

 

[jambaugh]

To: jambaugh,

...
By the way, that's a pretty disturbing "icon" you got there!

Don.

Yea! I was playing with time exposures. This one of me I call "Superposition of States".

 

[Hurkyl]

I've given a lot of lee-way thus far, but I see no reason to allow this thread to continue. You, Don Blazys, seem to be primarily interested in advertising how "revolutionary" your work is, and to be entirely uninterested in the standards mathematical proof -- therefore, this forum is an inappropriate place for your musings. Being obnoxious to your critics ("oops, I proved it again" ) doesn't help your case either.