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.

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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 seperate 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.

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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!

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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

Homework Helper
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.

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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 cancelled 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.

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CRGreathouse

Homework Helper
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

Homework Helper
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.

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Office_Shredder

Staff Emeritus
Gold Member
Line 4 isn't right

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

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

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 cancelled $$T$$'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 cancelled value" and to allow $$T=c$$.

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

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

Oops, I proved it again!

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 cancelled 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.

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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,

Don.

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