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

**Don Blazys** · Mar20-09, 04:46 AM

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

if

,
and

are co-prime, then

are not all greater than

.

Proof:

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

._________________________________________________ __________(1)

Factoring (1) results in:

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

$LaTeX Code: 1=\\left(\\frac{T}{T}\\right),$ where

._________________________________________________ (3)

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

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

and:

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

At this point we note that the definition of unity in (3) implies: $LaTeX Code: 1=\\left(\\frac{T}{T}\\right)=\\left(\\frac{c}{c}\\right ),$
which clearly means that

must be allowable.
We also note that the logarithms preventing

"cancel out" and therefore
cease to exist if and only if

in (4), and

in (5), which gives us both:

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

and:

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

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

__________________________________________________ _________(8)

and:

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

which proves not only the Beal Conjecture, but Fermat's Last Theorem
(which is only the special case where

) as well.

~~imag94~~ · Mar20-09, 10:36 AM

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~~ · Mar20-09, 10:47 AM

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** · Mar20-09, 10:56 AM

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 $LaTeX Code: \\frac{ln(T)}{ln(c)}-1$ 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** · Mar21-09, 12:52 AM

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).

must be allowable. Now, in order to allow

, we must follow this "two step" procedure in exactly this order.

(Step one) let

in equation (4) and

in equation (5).

This gives us both:

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

and:

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

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

This gives us both:

$LaTeX Code: T\\left(\\frac{c}{T}\\right)-\\left(\\frac{T}{T}\\right)b^y=\\left(\\frac{T}{T}\\righ t)a^x$

and:

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

Now and only now can we allow , 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

,
then we have either "division by zero", or the inability to allow

, both of which are unacceptable.

Don.

**Don Blazys** · Mar21-09, 04:48 AM

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:

which resulted in:

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

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 $LaTeX Code: 1=\\left(\\frac{T}{T}\\right),$
and substituted a newly discovered term which resulted in:

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

and:

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

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

must be allowable,
and that the only way to allow

is to first let

and

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 $LaTeX Code: 1=\\left(\\frac{T}{T}\\right),$ 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

while the factored case requires

in order to cancel out the logarithms and let

.)
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

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

implies

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

and that

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

and

respectively.
Then and only then can we allow

.

Surely you are not suggesting that

is prohibited ???!!!

Don.

**maze** · Mar21-09, 06:59 AM

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

$LaTeX Code: 0 = 1 \\cdot 0 = \\frac{0}{0} \\cdot 0 = \\frac{0 \\cdot 0}{0} = \\frac{0}{0} = 1$

**wsalem** · Mar21-09, 08:59 AM

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** · Mar21-09, 04:34 PM

To: maze,

Quoting maze:

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

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

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

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

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

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

themselves "cancel out" at

and

respectively, $LaTeX Code: \\frac{0}{0}=1$ 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

and

respectively, before letting

.

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:

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

and:

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

Notice that in order to allow

(which common sense tells us must be allowable),
we must first take the following two steps in exactly this order:

(Step one) Let

in (4) and

in (5). This gives us:

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

and:

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

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

$LaTeX Code: T\\left(\\frac{c}{T}\\right)-\\left(\\frac{T}{T}\\right)b^y=\\left(\\frac{T}{T}\\righ t)a^x$

and:

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

Now and only now can we let , which gives us:

$LaTeX Code: c\\left(\\frac{c}{c}\\right)-\\left(\\frac{c}{c}\\right)b^y=\\left(\\frac{c}{c}\\righ t)a^x$

and:

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

So you see, it is possible to allow

provided that we first
"cancel out" the logarithms at

and

respectively.
Oops I proved it again!

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

Don.

**CRGreathouse** · Mar21-09, 05:00 PM

Originally Posted by Don Blazys

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** · Mar21-09, 05:12 PM

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

and thereby prevent unity from being defined "in terms of itself" as: $LaTeX Code: 1=\\frac{1}{1}.$

As for your "assertion" that "

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** · Mar21-09, 06:58 PM

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

,
then

would have cancelled out initially.
Therefore, since my proof does not contain the "extra variable"

,
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

.

(By the way, the expression $LaTeX Code: 1=\\frac{T}{T}$ 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:

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** · Mar21-09, 09:30 PM

Originally Posted by Don Blazys

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** · Mar22-09, 01:59 AM

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:

results in:

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

and multiplying

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

results in:

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

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

.

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** · Mar22-09, 02:14 AM

Originally Posted by Don Blazys

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!

Originally Posted by Don Blazys

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.

Originally Posted by Don Blazys

In other words, factoring:

results in:

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

and multiplying

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

results in:

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

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
$LaTeX Code: c^{\\frac{z}{2}}+b^{\\frac{y}{2}}$ and $LaTeX Code: c^{\\frac z2}-b^{\\frac y2}.$
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 $LaTeX Code: c^{z/2}+b^{y/2}$ and $LaTeX Code: c^{z/2}-b^{y/2}$ are integers.

**Don Blazys** · Mar22-09, 05:12 PM

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 $LaTeX Code: c^{\\frac{z}{2}}+b^{\\frac{y}{2}}$ and $LaTeX Code: c^{\\frac z2}-b^{\\frac y2}.$

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 $LaTeX Code: c^{z/2}+b^{y/2}$ and $LaTeX Code: c^{z/2}-b^{y/2}$ are integers.

In both the factorable and unfactorable cases, if we assume that the terms are co-prime,
then we automatically assume that

are co-prime.
The individual terms that the factorable case implies are then:

$LaTeX Code: \\left(c^\\frac{z}{2}\\right), \\left(b^\\frac{y}{2}\\right)$ and

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

,

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

,

then we would properly do so in the unfactorable case:

.

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

Don.