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

[Don Blazys]

An Elementary Proof Of The Beal Conjecture And Fermat's Last Theorem.
By: Don Blazys.

The Beal Conjecture can be stated as follows:

For positive integers $$a,b,c,x,y,z,$$ if $$a^x+b^y=c^z$$,
and $$a,b,c$$ are co-prime, then $$x,y,z$$ are not all greater than $$2$$.

Proof:

Letting all variables herein represent positive integers, we form the equation:

$$c^z-b^y=a^x$$.___________________________________________________________(1)

Factoring (1) results in:

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

Here, it will be assumed that the terms in (1) and (2) are co-prime,
and that the only "common factor" they contain is the "trivial" unity,
which can not be defined in terms of itself, and must therefore be defined as:

$$1=\left(\frac{T}{T}\right),$$ where $$T>1$$._________________________________________________(3)

Re-stating (1) and (2) so that the "trivial common factor" $$1=\left(\frac{T}{T}\right)$$
and it's newly discovered logarithmic consequences are represented, we now have both:

$$T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x$$ ____________________________________(4)

and:

$$\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(\left(\frac{T}{T}\right)c^\frac{z}{2}-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x.$$ ___________(5)

At this point we note that the definition of unity in (3) implies: $$1=\left(\frac{T}{T}\right)=\left(\frac{c}{c}\right),$$
which clearly means that $$T=c$$ must be allowable.
We also note that the logarithms preventing $$T=c$$ "cancel out" and therefore
cease to exist if and only if $$z=1$$ in (4), and $$z=2$$ in (5), which gives us both:

$$T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x$$ ____________________________________________(6)

and:

$$\left(T\left(\frac{c}{T}\right)+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x.$$ _____________________(7)

$$T=c$$ is now clearly allowable, and simplifying (6) and (7) shows that
the original equations, as stated in (1) and (2), are now:

$$c-b^y=a^x,$$ ___________________________________________________________(8)

and:

$$\left(c+b^\frac{y}{2}\right)\left(c-b^\frac{y}{2}\right)=c^2-b^y=a^x,$$_________________________________________(9)

which proves not only the Beal Conjecture, but Fermat's Last Theorem
(which is only the special case where $$x=y=z$$) as well.

 

[imag94]

The proof seems ok, except in equation 4) and 5) ln c/T when T=c is indeterminate being infinity(division by zero)

_________________
Mathew Cherian

 

[imag94]

This proof is OK, but in equation 4) and 5) we can have c and T >2 which will give an escape route from depending on the square identity of a, b and c. For practical purpose even higher powers can work.

________________
Mathew Cherian

 

[wsalem]

You reached a very obvious contradiction.
In 8 and 9, you showed that c = c^2. This implies c = 1. But then c = a^x + b^y and by assumption a,b are positive integers, a contradiction.
You can be assured that you did something wrong!

Assuming that T=c, and dividing by $$\frac{ln(T)}{ln(c)}-1$$ is meaningless... Also claiming that "that this is meaningful if and only if z = 1" is incorrect. Actually since c turned out to be equal to 1, division by ln(c) is not even allowed.
c equals 1 and T=c is a contradiction, since T>1.
\frac{T}{T} = \frac{c}{c} implies T=c is certainly not true.

 

[Don Blazys]

To:imag94,


Quoting imag94:

The proof seems ok, except in equation 4) and 5) ln c/T when T=c is indeterminate being infinity(division by zero).

$$T=c$$ must be allowable. Now, in order to allow $$T=c$$, we must follow this "two step" procedure in exactly this order.

(Step one) let $$z=1$$ in equation (4) and $$z=2$$ in equation (5).

This gives us both:

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

and:

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

(Step two) "Cross out" or "cancel out" the expressions involving logarithms.

This gives us both:

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

and:

$$\left(T\left(\frac{c}{T}\right)+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x.$$

Now and only now can we allow $$T=c$$ , which common sense tells us must be allowable.

You see, by following this simple "two step" procedure, not only were we able to avoid
"divisions by zero" and "indeterminate forms", but we actually proved
both the Beal Conjecture and Fermat's Last Theorem, since we clearly showed that if $$z>2$$,
then we have either "division by zero", or the inability to allow $$T=c$$, both of which are unacceptable.

Don.

 

[Don Blazys]

To: wsalem,


Quoting wsalem:

You can be assured that you did something wrong!

I did nothing wrong! Let's go over it "step by step".

All I did was factor the equation:

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

which resulted in:

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

There's nothing "wrong" with that. It's "standard procedure" that you can find in any algebra textbook.

Then, in the above two equations, I merely multiplied each and every term by $$1=\left(\frac{T}{T}\right),$$
and substituted a newly discovered term which resulted in:

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

and:

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

There's nothing "wrong" with that either. Multiplying by unity does not change anything,
and substitutions involving identities are also perfectly legal.

Lastly, all I did was point out that $$T=c$$ must be allowable,
and that the only way to allow $$T=c$$ is to first let $$z=1$$ and $$z=2$$ respectively,
then cancel out or "cross out" the expressions involving logarithms so that they no longer exist.

Oops I proved it again!

You see, this is a very, very straightforward result!

Simply factor, multiply by $$1=\left(\frac{T}{T}\right),$$ substitute, and bingo!


quoting wsalem:

You reached a very obvious contradiction.
In 8 and 9, you showed that c = c^2. This implies c = 1. But then c = a^x + b^y and by assumption a,b are positive integers, a contradiction.

Actually since c turned out to be equal to 1, division by ln(c) is not even allowed.
c equals 1 and T=c is a contradiction, since T>1.

Equations (8) and (9) are the results of two separate cases.
("Not factorable" and "factorable" respectively.)
Each case requires it's own unique set of positive integers.
(This is quite obvious, since the unfactored case requires $$z=1$$ while the factored case requires $$z=2$$ in order to cancel out the logarithms and let $$T=c$$.)
Thus they constitute two different equations and two independent representations.

Your assertion that the above equations "imply" c=1 is therefore without foundation,
as are the rest of your "critiques".

For instance:


Quoting wsalem:

\frac{T}{T} = \frac{c}{c} implies T=c is certainly not true.

Where in my proof did I say that? What I said is that

$$1=\left(\frac{T}{T}\right)$$

implies

$$1=\left(\frac{T}{T}\right)=\left(\frac{c}{c}\right ),$$

and that $$T=c$$ must therefore be allowable in equations (4) and (5), which, in turn,
requires that we first "cancel out" or "cross out"
the expressions involving logarithms at $$z=1$$ and $$z=2$$ respectively.
Then and only then can we allow $$T=c$$.

Surely you are not suggesting that $$T=c$$ is prohibited ?!

Don.

 

[maze]

In an expression of the form 0/0, the zeros do not cancel. Consider this "proof" that 0=1:

$$0 = 1 \cdot 0 = \frac{0}{0} \cdot 0 = \frac{0 \cdot 0}{0} = \frac{0}{0} = 1$$

 

[wsalem]

Surely you are not suggesting that T=c is prohibited ?!

Reread my post, yes T=c must be prohibited, as well as T=1 (You can't write equation 4 without explicitly assuming this).
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!

 

[Don Blazys]

To: maze,


Quoting maze:

In an expression of the form 0/0, the zeros do not cancel. Consider this "proof" that 0=1:
$$0 = 1 \cdot 0 = \frac{0}{0} \cdot 0 = \frac{0 \cdot 0}{0} = \frac{0}{0} = 1$$

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

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

Moreover, in many cases, there are various ways by which we can determine

the meaning or value of an indeterminate form such as: $$\frac{0}{0}.$$

In the case of my proof, it is sufficient to note that since the logarithms that engendered $$\frac{0}{0}$$

themselves "cancel out" at $$z=1$$ and $$z=2$$ respectively, $$\frac{0}{0}=1$$ is a "logically consistent" interpretation.

Besides, the only reason that you encountered that pesky "indeterminate form" is because
you forgot to "cancel out" the expressions involving logarithms at $$z=1$$ and $$z=2$$ respectively, before letting $$T=c$$.

As I already demonstrated in post#5, if we are sufficiently clever,
then we can actually avoid those pesky "indeterminate forms" altogether!

For your convenience, I will repeat that demonstration now. Here then is:

A demonstration that "indeterminate forms" in Don's proof are avoidable.

Consider equations (4) and (5) in the proof. They are:

$$T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x$$____________________________________________(4)

and:

$$\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(\left(\frac{T}{T}\right)c^\frac{z}{2}-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x.$$____________________(5)

Notice that in order to allow $$T=c$$ (which common sense tells us must be allowable),
we must first take the following two steps in exactly this order:

(Step one) Let $$z=1$$ in (4) and $$z=2$$ in (5). This gives us:

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

and:

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

(Step two) Cancel out or "cross out" the expressions involving logarithms
so that they no longer exist. This gives us:

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

and:

$$\left(T\left(\frac{c}{T}\right)+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x.$$

Now and only now can we let $$T=c$$, which gives us:

$$c\left(\frac{c}{c}\right)-\left(\frac{c}{c}\right)b^y=\left(\frac{c}{c}\right)a^x$$

and:

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

So you see, it is possible to allow $$T=c$$ provided that we first
"cancel out" the logarithms at $$z=1$$ and $$z=2$$ respectively.
Oops I proved it again!

Most importantly, notice that no pesky "indeterminate forms" were ever encountered!
Isn' that great?!

Don.

 

[CRGreathouse]

There's nothing "wrong" with that. It's "standard procedure" that you can find in any algebra textbook.

Right, there's nothing wrong with that step. But your following assumptions that the terms are (1) integers, and (2) coprime are not justified.

 

[Don Blazys]

To: wsalem,

Quoting wsalem:

Reread my post, yes T=c must be prohibited, as well as T=1 (You can't write equation 4 without explicitly assuming this).
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!

The proof works precisely because
"Blazys terms" (the ones involving logarithms) preclude $$T=1$$
and thereby prevent unity from being defined "in terms of itself" as: $$1=\frac{1}{1}.$$

As for your "assertion" that "$$T=c$$ must be prohibited",
well, in my previous post (post #9), I demonstrated quite conclusively
that all such "assertions" are utterly vaccuous and that you are therefore mistaken.

Don.

 

[Don Blazys]

To: CRGreathouse,

Quoting CRGreathouse:

Right, there's nothing wrong with that step. But your following assumptions that the terms are (1) integers, and (2) coprime are not justified.

The assumptions of both co-primality and positive integer variables are indeed "justified".

The justification for the assumption of co-primality is as follows:
If the terms did contain some common factor $$N$$,
then $$N$$ would have canceled out initially.
Therefore, since my proof does not contain the "extra variable" $$N$$,
it is logical (and therefore justifiable) to assume that the terms are co-prime.
In other words, if we wanted our terms to be not co-prime,
then we would simply multiply them by $$N$$.

(By the way, the expression $$1=\frac{T}{T}$$ can,
in some versions of my proof, be viewed as the "cancelled common factor").


The justification for the assumption of positive integer vaviables is somewhat similar.
If we wanted to include, say, negative integers,
then we could always represent them as: $$(-a), (-b), (-c)$$ etc.
Therefore, since zero and the positive integers are the only numbers that
do not require some operation as part of their representation,
it is logical (and therefore justifiable) to assume that our variables represent positive integers.

Please note that the above "justifications" are largely viewed as "inherent in number theory",
and most books on "number theory", "pure math", and "recreational math" routinely
assume both "co-primality" and "positive integer variables", often without any further explanation,
which is, in my opinion, not very kind to those who are just beginning to study this fine subject.

Don.

 

[CRGreathouse]

The assumptions of both co-primality and positive integer variables are indeed "justified".

You haven't even defended the integrality of the terms! If they're not integers, then "being coprime" is meaningless. What about the (likely) case that at least one of y and z is odd?

 

[Don Blazys]

To: CRGreathouse,

Quoting CRGreathouse:

You haven't even defended the integrality of the terms! If they're not integers, then "being coprime" is meaningless. What about the (likely) case that at least one of y and z is odd?

Just because I'm "old" doesn't mean that I am so "wise" as to be able to read peoples minds!
You asked me to "justify" the assumptions of co-primality and positive integer variables,
and that's exactly what I "delivered"! This is a new question, and a good one!

As for "defending the integrality of the terms", well, since the variables are "justifiably"
representative of positive integers only, it immediately follows that the terms will also be
representative of positive integers only.

You see, when we factor an equation, we implement a system of "proper values" for a
particular type of equation, but that in no way compromises the integrality of the terms,
even when "improper values" are plugged into the variables.

In other words, factoring:

$$c^z-b^y=a^x$$ results in:

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

and multiplying

$$\left(c^\frac{z}{2}+b^\frac{y}{2}\right)$$ by $$\left(c^\frac{z}{2}-b^\frac{y}{2}\right)$$

results in:

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

and eliminating the outermost parenthesis brings us back to our original "unfactored" equation:

$$c^z-b^y=a^x$$.

So even if we "improperly" plug in "non-square" values into the above factorization,
in the end, we will still have terms that are integers.

I hope this clarifies the fact that positive integer variables guarantee the "integrality" of the terms.

My proof relies solely on "Blazys terms" (the terms involving logarithms),
and "works" perfectly regardless of how we factor, because the "extra information"
they contain is sufficient to prevent "improper values" from being plugged into the variables.

Don.

 

[CRGreathouse]

Just because I'm "old" doesn't mean that I am so "wise" as to be able to read peoples minds!
You asked me to "justify" the assumptions of co-primality and positive integer variables,
and that's exactly what I "delivered"! This is a new question, and a good one!

Actually, I asked you to justify the assumption that the terms were "(1) integers, and (2) coprime", so to me this isn't new at all -- it was the first thing I asked!

As for "defending the integrality of the terms", well, since the variables are "justifiably"
representative of positive integers only, it immediately follows that the terms will also be
representative of positive integers only.

Integers aren't closed under square roots. 5^7 is an integer, but 5^(7/2) is not.

In other words, factoring:

$$c^z-b^y=a^x$$ results in:

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

and multiplying

$$\left(c^\frac{z}{2}+b^\frac{y}{2}\right)$$ by $$\left(c^\frac{z}{2}-b^\frac{y}{2}\right)$$

results in:

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

I don't think we're communicating here. You wrote "the terms in (1) and (2) are co-prime", and by "the terms" I assumed you meant
$$c^{\frac{z}{2}}+b^{\frac{y}{2}}$$ and $$c^{\frac z2}-b^{\frac y2}.$$
If not, what did you mean? If so, it's not enough to show that the product is a^x (no one disagrees with that), but that $c^{z/2}+b^{y/2}$ and $c^{z/2}-b^{y/2}$ are integers.

 

[Don Blazys]

To: CRGreathouse,

Quoting CRGreathouse:

I don't think we're communicating here. You wrote "the terms in (1) and (2) are co-prime",
and by "the terms" I assumed you meant $$c^{\frac{z}{2}}+b^{\frac{y}{2}}$$ and $$c^{\frac z2}-b^{\frac y2}.$$

If not, what did you mean? If so, it's not enough to show that the product is a^x (no one disagrees with that),
but that $c^{z/2}+b^{y/2}$ and $c^{z/2}-b^{y/2}$ are integers.

In both the factorable and unfactorable cases, if we assume that the terms are co-prime,
then we automatically assume that $$a, b, c$$ are co-prime.
The individual terms that the factorable case implies are then:

$$\left(c^\frac{z}{2}\right), \left(b^\frac{y}{2}\right)$$ and $$a^x$$

As for showing that $\left(c^{z/2}+b^{y/2}\right)$ and $\left(c^{z/2}-b^{y/2}\right)$ are integers, well, if $$c=5, z=2, b=2, y=4$$,

then both of the expressions within the parenthesis are indeed integers,

as are the individual terms themselves.

However, if we wanted to assign values such as $$c=3, z=3, b=2, y=3$$,

then we would properly do so in the unfactorable case: $$c^z-b^y=a^x$$.

It can then be shown that between both cases, all possibilities are covered.

Don.

 

[Office_Shredder]

Line 4 isn't right

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


should read

$$\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac {\frac{z}{2}\ln(c)}{\ln(T)}-z/2}{\frac{\ln(c)}{\ln(T)}-1}\right)}+\left(\frac{T}{T}\right)b^\frac{y}{2}\r ight)\left(\left(\frac{T}{T}\right)c^\frac{z}{2}-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left (\frac{T}{T}\right)a^x.$$


You didn't distribute the z/2 properly on the c term

 

[Don Blazys]

To: Office Shredder,

Equation (4) in my proof is:

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

and equation (5) is:

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

so I'm pretty sure that you meant equation (5).

Now, you will not find anything like this equation in any textbook, journal or magazine
because it is only ten years old and has not yet been published.

It's properties are quite different from what most teachers, students
and even professional mathematicians are used to.

Notice that we can't just "cancel out" or "cross out" the canceled $$T$$'s,
unless we first let z=1 in equation (4) and z=2 in equation (5),
which, in and of itself, proves both the Beal Conjecture and Fermat's Last Theorem,
because both common sense and logic tell us that
it must be possible to "cancel out a canceled value" and to allow $$T=c$$.

After all, logarithms themselves would be badly flawed if both "cancelling out a canceled value"
and allowing $$T=c$$ were "impossible" !

Thus, the above two equations and accompanying observations are the entire proof.

Oops, I proved it again!

As for your assertion that:

Quoting Office Shredder:

You didn't distribute the z/2 properly on the c term.

That's not true. Here's why.

$$\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}\right)^2= \left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln(c)}{\ln(T)}-\frac{T}{T}}{\frac{\ln(c)}{\ln(T)}-1}\right)}\right)^2= T^2\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln(c)}{\ln(T)}-2}{\frac{\ln(c)}{\ln(T)}-1}\right)}= T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}$$

Notice that when we eliminate the outermost parenthesis in the first two terms,
we get two different possibilities, depending on whether or not
we involve the canceled value $$T$$.

Also notice that the first three terms require $$z=2$$ in order to "cancel out"
the expressions involving logarithms while the last term requires $$z=1$$.

Pretty amazing, isn't it ?!

Compare this to:

$$\left(\left(\frac{T}{T}\right)c^\frac{z}{2}\right)^2= \left(\frac{T}{T}\right)^2c^z= c^z,$$

where eliminating the outermost parenthesis

causes us to "lose track" of the fact that a factorization ever occured!

Believe it or not, once you get used to working with "Blazys terms",
they become quite easy... and open up a whole new realm of possibilities.

Don.

 

[csprof2000]

...
I'm pretty sure faulty logic is still faulty logic even if you include an arbitrary number of exclamation points. Unless, of course, you include an uncountably infinite number of exclamation points, in which case people couldn't prove your argument was flawed using induction.

 

[Don Blazys]

To: csprof2000,

Quoting csprof2000

I'm pretty sure faulty logic is still faulty logic even if you include an arbitrary number of exclamation points. Unless, of course, you include an uncountably infinite number of exclamation points, in which case people couldn't prove your argument was flawed using induction.

The exclamation marks convey both my exitement and enthusiasm for the result.

If you have no intelligent questions and can neither confirm or refute the result,
then please leave my thread alone.

Don.

 

[csprof2000]

The equation in line 8 is clearly equivalent to the equation in line 1 iff z = 1, since all numbers are positive integers. This is so clearly obvious that I don't even see the need to explain why.

Nobody ever said that z = 1 in the problem description. In fact, let a=3, b=4, c=5, and x=y=z=2. Then the equation in line 1 holds, but the one in line 8 does not. Nobody ever said z = 1.

Your logic between lines 1 and 8 is so tortured and amateurish that I won't dignify it with any more detailed analysis. If you're so sure that you're right, submit your proof to a journal for publication. Make sure to post the reply you get, if you get one... it should be good for a few laughs. If you get published, by all means, this is a great discovery, since you have essentially proved that it can be assumed that wherever you see the variable "z" in any equation you can replace it with "1". Awesome.

I have a few proofs of Fermat's last theorem in the special case where x=y=z=3 and a,b,and c can also be assumed. What's more, mine are coherent.

 

[Don Blazys]

To: csprof2000

Quoting csprof2000:

I won't dignify it with any more detailed analysis.

Your "analysis" was indeed "detailed".

Thanks for the astute observations.

Don.

 

[csprof2000]

Anytime. Hey, if you have any more "proofs" that involve factoring out a 1 from terms to show that z=1 for all z in the Real Numbers, let me know about it! Nice method. Wow.

Hey, I bet you could prove the Riemann hypothesis like that, too. Heck, you could probably show that P = NP into the deal. I'm honored to have read this enlightening thread.

 

[CRGreathouse]

As for showing that $\left(c^{z/2}+b^{y/2}\right)$ and $\left(c^{z/2}-b^{y/2}\right)$ are integers, well, if $$c=5, z=2, b=2, y=4$$,

then both of the expressions within the parenthesis are indeed integers,

as are the individual terms themselves.

However, if we wanted to assign values such as $$c=3, z=3, b=2, y=3$$,

then we would properly do so in the unfactorable case: $$c^z-b^y=a^x$$.

It can then be shown that between both cases, all possibilities are covered.

How come that doesn't appear in your proof?

 

[Don Blazys]

To: CRGreathouse,


quoting CRGreathouse:

How come that doesn't appear in your proof?

You are absolutely right. That is indeed a "key point"
and its omission was the reason for our initial failure to communicate.

Please accept my apology as I did not leave it out to create confusion.
(There is plenty of that already!) It was just an oversight...
an "honest mistake" on my part.

Also, please know that you have my gratitude for making
the first truly constructive suggestion that I got in quite a while.

I will definitely edit my proof so that it includes this "key point",
and would welcome any suggestions as to where in the proof it should go
and how it can best be stated using the least number of words.

Don.

 

[Hurkyl]

and how it can best be stated using the least number of words.

No. Worry about making correct, precise, and clear statements. Wait until you have actually confirmed that you have a valid proof before you even begin worrying about the "best" way to present it. (Or how to express it in the fewest words, if for some strange reason you care about that)

And tone the belligerence way down.

 

[maze]

I will admit that you might be able to get around the division by zero issue by understanding (4) and (5) in the sense of limits, but there is still a major structural problem to the proof.

If what you have done in (1), (2), and (3) is correct, then (4) and (5) must both be satisfied simultaneously. However, like you say, (4) is only satisfied for z=1 and (5) is only satisfied for z=2, but there is no value of z that satisfies both. Thus it seems you have shown there are no solutions at all.

 

[Don Blazys]

To: Hurkyl,

Quoting Hurkyl:

No. Worry about making correct, precise, and clear statements. Wait until you have actually confirmed that you have a valid proof before you even begin worrying about the "best" way to present it. (Or how to express it in the fewest words, if for some strange reason you care about that)

And tone the belligerence way down.

Thanks for the sound advice. I am by no means perfect,
but pease be assured that I am doing my very best to avoid
making any statements that are either wrong or belligerent.

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.

Here's what I got so far.

Wrong Statements Made By Others

You didn't distribute the z/2 properly on the c term...___________(1)
Easy to check.

Reread my post, yes T=c must be prohibited..._________________(2)
Demonstrated wrong in posts #5 and #9.

In an expression of the form 0/0, the zeros do not cancel...______(3)
Demonstrated avoidable and therefore
equal to unity in posts #5 and #9.

In 8 and 9, you showed that c = c^2..._______________________(4)
(8) and (9) are "seperate cases".
(Unfactorable and factorable.)

\frac{T}{T} = \frac{c}{c} implies T=c is certainly not true..._____(5)
I never said that. I said 1=T/T implies 1=c/c

T=c is indeterminate being infinity(division by zero)...____________(6)
Demonstrated avoidable and equal to unity
in posts #5 and #9.

Belligerent Statements Made By Others:

Your logic is tortured and amateurish...________________________(1)

I won't dignify it with any more detailed analysis..._______________(2)

It should be good for a few laughs...__________________________(3)

Hey, if you have any more "proofs"...__________________________(4)

Hey, I bet you could prove ...________________________________(5)

Heck, you could probably show..._____________________________(6)

I'm pretty sure faulty logic is still faulty logic...__________________(7)


Wrong statements made by Don.

Omitted explanation on "Proper" variable assignments...____________(1)

Belligerent Statements Made By Don.

__________________________________________________________(0)



As you can see, I have endured and absorbed many unprovoked insults
without even once "retaliating", and made only one "honest mistake".

Most importantly, I also have the courage to admit my mistakes.

Don.

 

[Don Blazys]

To: maze,

Quoting maze:

I will admit that you might be able to get around the division by zero issue by understanding (4) and (5) in the sense of limits, but there is still a major structural problem to the proof.

If what you have done in (1), (2), and (3) is correct, then (4) and (5) must both be satisfied simultaneously. However, like you say, (4) is only satisfied for z=1 and (5) is only satisfied for z=2, but there is no value of z that satisfies both. Thus it seems you have shown there are no solutions at all.

If that was indeed the case, then the fault would lie in the properties of logarithms.
We would then have to ban logarithms, wouldn't we?

However, we are dealing with positive integers, and maintaining integrality necessitates that
the variables in (1) and (2) have unique and separate domains.

Thus, by definition, (4) and (5) must be satisfied seperately, not simultaneously.
In other words, they are two separate cases, which, when taken together, cover all possibilities.

Please read the dialogue between CRGreathouse and myself.

We already covered this issue and I will soon revise my proof to include that information.

Don.

 

[csprof2000]

I think I just realized you're a regular, Don, taking us all around the block for a nice E-joke. It would have been funnier if your username had been "TheGrandKermoctopus".

 

[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.