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

[Don Blazys]

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

By: Don Blazys

 

The Beal Conjecture can be stated as follows:

 

For positive integers: [math]a, b, c, x, y,[/math] and [math]z[/math], if [math]a^x+b^y=c^z[/math],

and [math]a, b,[/math] and [math]c[/math] are co-prime, then [math]x, y, [/math] and [math]z[/math] are not all greater than [math]2[/math].

 

Proof:

 

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

 

[math]c^z-b^y=a^x[/math]. _______________________________________________________________(1)

 

Factoring (1) results in:

 

[math]\left(c^\frac{z}{2}+b^\frac{y}{2}\right)\left(c^\frac{z}{2}-b^\frac{y}{2}\right)=a^x[/math]. _________________________________________________(2)

 

where (1) and (2) are now different, distinct and seperate cases,

which can not be considered or satisfied simultaneously because

the integrality of the terms in (2)

requires that both [math]y[/math] and [math]z[/math] be evenly divisible by [math]2[/math].

 

Here, it will be assumed that [math]a, b[/math] and [math]c[/math] are co-prime,

so that the only common factor possible is the "trivial" unity,

which can not be defined in terms of itself, and must therefore be defined as:

 

[math]1=\left(\frac{T}{T}\right)[/math], where [math]T>1[/math]. _____________________________________________________(3)

 

Re-stating (1) and (2) so that the "trivial common factor" [math]1=\frac{T}{T}[/math] and its

newly discovered logarithmic consequences are represented, we now have both:

 

[math]T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x[/math] _____________________________________(4)

 

and:

 

[math]\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln©}{\ln(T)}-1}{\frac{\ln©}{\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[/math]._________(5)

 

where it is clear that the expressions involving logarithms "cancel out"

if and only if [math]z=1[/math] in (4), and [math]z=2[/math] in (5), which gives us both:

 

[math]T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x [/math], ______________________________________________(6)

 

and:

 

[math] \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[/math] . ____________________(7)

 

Here we note that the definition of unity in (3) implies: [math]1=\left(\frac{T}{T}\right)=\left(\frac{c}{c}\right)[/math].

 

Therefore, [math]T=c[/math] must be possible,

because substituting [math]\frac{c}{c}[/math] for [math]\frac{T}{T}[/math] is exactly the same as substituting [math]1[/math] for [math]1[/math],

and can not possibly effect either the outcome or the integrity of the result.

 

Thus we find that both the Beal Conjecture and Fermat's Last Theorem

(which is only the "special case" where [math]x=y=z[/math]),

must be true, since [math]T=c[/math] is possible only in (6) and (7),

where [math]z=1[/math] in (4), and [math]z=2[/math] in (5)

caused the expressions involving logarithms to "cancel out".

 

In other words, the impossibility of substituting [math]\frac{c}{c}=1[/math] for [math]1=\frac{T}{T}[/math] is a contradiction,

and since that contradiction occurs only when the assumption is made that

[math]x,y,z[/math] can all be greater than [math]2[/math] when [math]a^x+b^y=c^z[/math] and [math]a,b,c[/math] are co-prime,

that assumption must be wrong and the conjecture has been proved.

 

Don Blazys

[Don Blazys]

Humbly presented as a gift to mankind by yours truly.

 

Don.

[CraigD]

The requirement that

[math]T=c[/math] must be allowable

results in (4) simplifying to

 

[math]\left(\frac{T}{T}\right)a^x + \left(\frac{T}{T}\right)b^y = T\left(\frac{c}{T}\right)^{\left(\frac{z-1}{0}\right)}[/math]

 

which is indeterminate.

 

Even if one accepts the indeterminate value in the proof, the proof has another serious problem: without an unstated assumption, it can be used to prove a conjecture that can be show false by example, as follows.

 

Replacing “2” in (2) with “3” gives

 

[math](a^\frac{x}{3})^3 + (b^\frac{y}{3})^3 = (c^\frac{z}{3})^3[/math]

 

which lead to the conclusion that

 

[math]a^z +b^z = c^z[/math]

 

is true if and only if z=1 or z=3. However, it’s know by example that [math]a^z +b^z = c^z[/math] is true when z=2 for any Pythagorean triple {a,b,c}, eg: [math]3^2 +4^2 = 5^2[/math].

 

The same problem appears if any value is used in place of “2” in (2).

 

Hence, even if all other parts of the proof are accepted, the proof is consistent with the Pythagorian theorem only if makes the unstated assumption to use “2” in (2). Doing so requires that the proof assume what it is to prove. A proof that assumes what it is to prove utilizes circular reasoning, and is invalid.

[Don Blazys]

To: Craig D,

 

In equation (4), if [math]z=1[/math] then,

 

Quoting Craig D:

[math]\left(\frac{T}{T}\right)a^x + \left(\frac{T}{T}\right)b^y = T\left(\frac{c}{T}\right)^{\left(\frac{z-1}{0}\right)}

[/math]

 

is indeterminate.

 

Not necessarily!

If we are sufficiently clever, then we can actually avoid

those pesky "indeterminate forms" altogether!

Now, let's take another look at equation (4):

 

[math]\left(\frac{T}{T}\right)a^x + \left(\frac{T}{T}\right)b^y = \left(\frac{T}{T}\right)c^z = T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}[/math]

 

Notice that in order to allow [math]T=c[/math] (which must be allowable),

we must first take the following two steps in exactly this order:

 

(Step 1) Let [math]z=1[/math]. This gives us:

 

[math]\left(\frac{T}{T}\right)a^x + \left(\frac{T}{T}\right)b^y = \left(\frac{T}{T}\right)c = T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}[/math]

 

(Step 2) "Cancel out" the expressions involving logarithms so that they no longer exist.

This gives us:

 

[math]\left(\frac{T}{T}\right)a^x + \left(\frac{T}{T}\right)b^y = \left(\frac{T}{T}\right)c = T\left(\frac{c}{T}\right)[/math]

 

Now and only now can we let [math]T=c[/math], which gives us:

 

[math]\left(\frac{c}{c}\right)a^x + \left(\frac{c}{c}\right)b^y = \left(\frac{c}{c}\right)c = c\left(\frac{c}{c}\right)=a^x+b^y=c[/math]

 

Most importantly, notice that no pesky "indeterminate form" was ever encountered!

Isn't that great?!

 

Quoting Craig D:

Even if one accepts the indeterminate value in the proof...

 

As I just demonstrated, if we do the algebra correctly and "cancel out"

the expressions involving logarithms at [math]z=1[/math] before we let [math]T=c[/math],

then there simply is no pesky "indeterminate value" to "accept" or even speak of !

My proof, in post #1, actually shows that the "indeterminate value" that you encountered

because you failed to "cancel out" the expressions involving logarithms at [math]z=1[/math]

is nothing more than "dust in the wind" !

In actuality, it got "cancelled out" before it even had a chance to exist!

 

Quoting Craig D:

Replacing “2” in (2) with “3” gives

 

[math](a^\frac{x}{3})^3 + (b^\frac{y}{3})^3 = (c^\frac{z}{3})^3[/math]

 

which leads to the conclusion that:

 

[math]a^x+b^y=c^z[/math]

 

is true if and only if z=1 or z=3.

However, it’s know by example that is true when z=2 for any Pythagorean triple {a,b,c},

eg: [math]3^2+4^2=5^2[/math].

 

The same problem appears if any value is used in place of “2” in (2).

 

Hence, even if all other parts of the proof are accepted,

the proof is consistent with the Pythagorian theorem only if it

makes the unstated assumption to use “2” in (2).

Doing so requires that the proof assume what it is to prove.

A proof that assumes what it is to prove utilizes circular reasoning, and is invalid.

 

Holy "catch 22"!

 

Actually, the [math]2[/math]'s in (2) can be shown to occur as a direct result of factoring both:

 

[math]c^z-b^y=a^x[/math] and [math]c^z-a^x=b^y[/math].

 

Thus, they are not some "unstated assumption" but rather a consequence of factoring!

 

In other words, the [math]2[/math]'s in (2) exist because "Pythagorean triples" exist.

That's not an "unstated assumption" but a fact.

Facts are allowable. In fact, proofs require facts.

 

Thus, "circular reasoning" is not involved because the proof does not

"assume what it is to prove", which is that there must exist a restriction

on one of the variables if the terms are co-prime.

Note that [math]2[/math] is not a variable and that the actual restriction is not brought about

by the [math]2[/math]'s in (2), but rather by the introduction of logarithms in (4) and (5).

 

Most importantly, as you yourself pointed out,

we simply can't replace the [math]2[/math]'s in (2) with [math]3[/math]'s

without compromising "Pythagorean triples".

That is exactly why this result is utterly irrefutable!

 

Don.

[Tormod]

(a bunch of dancing smilies)

 

Don, I am one (among many) who don't understand mathematical notation too well so I follow threads like this by reading the claims and opposing arguments. Your use of dancing smilies makes it difficult to take your post seriously!

 

Just an opiniated voice of concern from me.

[CraigD]

For the sake of readability, let’s speak of a proof of the more widely known Fermat’s last theorem rather than Beal’s conjecture.

In other words, the [math]2[/math]'s in (2) exist because "Pythagorean triples" exist.

That's not an "unstated assumption" but a fact.

It’s not unstated now, but it was in post #1.

Thus, "circular reasoning" is not involved because the proof does not

"assume what it is to prove", which is that there must exist a restriction

on one of the variables if the terms are co-prime.

Two questions:

 

1) If there was known to exist a triple of integers [math]a,b,c > 1[/math] such that [math]a^3+b^3=c^3[/math], would the choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) be permitted?

 

2) How do you know there does not exist such a triple?

[Don Blazys]

To: Tormod,

 

Quoting Tormod:

Your use of dancing smilies makes it difficult to take your post seriously!

 

You are right.

 

This is a serious topic.

 

My attempt at trying to make it "more entertaining" by using "smilies"

was poorly concieved and more of a distraction than anything else.

Thanks for bringing it to my attention.

 

Those smilies are gone now.

 

I edited them out.

 

Don.

[Don Blazys]

To: Craig D,

 

Quoting Craig D:

Two questions:

 

1) If there was known to exist a triple of integers such that [math]a^3+b^3=c^3[/math],

would the choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) be permitted?

 

2) How do you know there does not exist such a triple?

 

Exellent questions.

 

Now, let's consider the equation:

 

[math]\left(\left(\frac{T}{T}\right)a^\frac{x}{q}\right)^q+\left(\left(\frac{T}{T}\right)b^\frac{y}{q}\right)^q=\left(\left(\frac{T}{T}\right)c^\frac{z}{q}\right)^q=\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{q}\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}\right)^q[/math] _______(8.1)

 

and ask ourselves how we should eliminate the outermost parenthesis.

 

Surprisingly enough, it turns out there are two equally valid ways to do that.

One way is to "disregard" the cancelled [math]T[/math]'s in (8.1) and write:

 

[math]\left(\frac{T}{T}\right)a^x+\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)c^z=T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}[/math], __________________________(8.2)

 

and the other way is to "involve" the cancelled [math]T[/math]'s in (8.1) and write:

 

[math]\left(\frac{T}{T}\right)^qa^x+\left(\frac{T}{T}\right)^qb^y=\left(\frac{T}{T}\right)^qc^z=T^q\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln©}{\ln(T)}-q}{\frac{\ln©}{\ln(T)}-1}\right)}[/math]. ______________________(8.3)

 

Now, let's list some facts about the above three equations.

 

fact #1: [math]q[/math] is a particular variable that represents

the "degree" to which equation (8.1) can be "factored".

Note that no assumption is being made as to the

value of [math]q[/math] and that [math]T=c[/math] must be allowable and requires that [math]z=q[/math].

 

fact #2: Since [math]T=c[/math] must be allowable,

equations (8.2) and (8.3) are both absolutely exclusive and mutually exclusive.

However, if it weren't for the terms involving logarithms,

then we would have absolutely no way of knowing that!

Moreover, if it weren't for the terms involving logarithms,

then we would simply lose both the notion of factorability,

and the concept of cancelled common factors!

(Notice that in both (8.2) and (8.3) the terms that

do not involve logarithms are virtually identical

and do not intrinsically retain any information whatsoever

regarding factorization or cancelled common factors.)

 

fact #3: A particular variable can not simultaneously represent

two or more different values in the same equation.

Thus, since eliminating the outermost parenthesis in equation (8.1)

resulted in only two different equations,

there can be only two different possible values for [math]q[/math],

and equation (8.2) shows that one of those values is unity,

while "Pythagorean triples" necessitate that the value of [math]q[/math] in (8.3) is [math]2[/math].

(In other words, there are only two ways to eliminate

the outermost parenthesis in equation (8.1) which means that

there is no "third way" that will give us yet another equation

and yet another possible value for [math]q[/math].)

 

fact #4: If [math]q>2[/math] was possible, (it's not),

then only the terms not involving logarithms would survive.

The terms involving logarithms would contradict

the existence of either "Pythagorean triples" or the [math]q>2[/math] equation,

and we would then be faced with the impossible task of having to determine

exactly why the properties of logarithms are not consistent with empirical facts,

and would ultimately have to eliminate the notion of logarithms from mathematics altogether!

Wouldn't that be both horrible and tragic!?

 

Thus, the answer to your first question:

 

Quoting Craig D:

1) If there was known to exist a triple of integers such that [math]a^3+b^3=c^3[/math],

would the choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) be permitted?

 

is as follows:

 

The choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) would then be permitted,

but we would then have to eliminate the logarithms in (4) and (5),

not to mention the rest of mathematics!

 

As for the second question,

 

Quoting Craig D:

2) How do you know there does not exist such a triple?

 

Well, for the longest time, we didn't know!

However, if we accept the relatively recent Wiles proof of FLT,

then we can say that we "know", and pretend that we "understand",

even though the vast majority of us will never really understand it!

Thus, for those of us who are not part of the "mathematical elite",

if we simply realize that the properties of logarithms are perfectly valid,

and, if used correctly, will never ever involve us in an inconsistency with the rest of mathematics,

then the proof at the beginning of this thread is, in my humble opinion,

more than sufficient to convince the "rest of us", that FLT is indeed true,

and that such a "triple" can't possibly exist.

 

I understand exactly what you mean by an "unstated assumption",

but if an "assumption" is duplicated by a "fact",

then the "fact aspect" of the "fact/assumption"

must supersede and take both priority and precedence

over the "assumption aspect" of the "fact/assumption",

and the "fact/assumption" must therefore be allowed

at the very beginning of the proof, where the data is first collected,

and facts are first gathered.

You see, this proof is not so much an "argument" as it is a "straightforward result".

In fact, it's almost "axiomatic" in that all we really did was preclude [math]T=1[/math].

(To my knowledge, this is the first time that such a preclusion was ever made using logarithms.)

The rest just follows!

 

I really do appreciate you providing me with the very astute observation that

the [math]2[/math]'s in (2) might misakenly be construed as an "unstated assumption",

and can, if you still think it is necessary, re-write the proof so that it

begins with the more general equations (8.1), (8.2) and (8.3) from this post,

then makes the argument for [math]q=2[/math] towards the end.

 

Either way, the proof is quite valid, and works perfectly.

 

Don.

[Don Blazys]

Okay.

 

So there we have it.

 

There are no "indeterminate forms",

 

and factoring is factual and does not constitute an "unstated assumption".

 

So far, all attempts (including my own) at finding a "fatal flaw" in this proof have failed .

 

So perhaps it's time that we own up to the fact that logarithms don't lie,

and perhaps admit that this result may be both true and correct.

 

Let's face it.

 

When a proof is this simple,

then there is almost no place for a "fatal flaw" to hide!

 

Thus, (at least to my mind) this proof may be valid.

 

The real question is now this,

 

Is this "proof" a hitherto undiscovered property of logarithms?

 

Don.

[CraigD]

In an attempt to explain to Don why his attempted proof of Fermat’s Last Theorem is wrong, I pointed out in post #3 that it’s necessary to assume in step 2 that [math]z=2[/math], not [math]z=3[/math] or any greater integer value, that this assumption requires accepting Fermat’s Last Theorem ([math]a^z+b^z=c^z[/math] has integer solution [math]a,b,c[/math] greater than 1 only for [math]z=1,2[/math]), and that assuming what you are proving is not valid because it is circular reasoning.

 

To this Don replied:

Holy "catch 22"!

 

Actually, the [math]2[/math]'s in (2) can be shown to occur as a direct result of factoring both:

 

[math]c^z-b^y=a^x[/math] and [math]c^z-a^x=b^y[/math].

 

Thus, they are not some "unstated assumption" but rather a consequence of factoring!

 

In other words, the [math]2[/math]'s in (2) exist because "Pythagorean triples" exist.

That's not an "unstated assumption" but a fact.

Facts are allowable. In fact, proofs require facts.

 

Thus, "circular reasoning" is not involved because the proof does not

"assume what it is to prove"

In an attempt to lead Don to discover for himself my objection, I asked:

1) If there was known to exist a triple of integers [math]a,b,c > 1[/math] such that [math]a^3+b^3=c^3[/math], would the choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) be permitted?

 

2) How do you know there does not exist such a triple?

Don answered the first question with a wordy “yes”, the second:

Well, for the longest time, we didn't know!

However, if we accept the relatively recent Wiles proof of FLT,

then we can say that we "know", and pretend that we "understand",

even though the vast majority of us will never really understand it!

Thus, for those of us who are not part of the "mathematical elite",

if we simply realize that the properties of logarithms are perfectly valid,

and, if used correctly, will never ever involve us in an inconsistency with the rest of mathematics,

then the proof at the beginning of this thread is, in my humble opinion,

more than sufficient to convince the "rest of us", that FLT is indeed true,

and that such a "triple" can't possibly exist.

In short, Don asserts that Fermat’s Last Theorem is true because of Wiles’s proof of it.

 

Assuming a proposition true, even if it is proved by another proof, in order to prove it, is circular reasoning, and not valid. Don’s proof is therefore incorrect.

[Don Blazys]

The following are reasons for why I am quite certain that

Craig D's "unstated assumption objection" is without foundation.

 

Quoting Craig D:

A summary of the discussion of one reason why Don’s proof is wrong.

 

In an attempt to explain to Don why his attempted proof of Fermat’s Last Theorem is wrong, I pointed out in post #3 that it’s necessary to assume in step 2 that [math]z=2[/math], not [math]z=3[/math] or any greater integer value, that this assumption requires accepting Fermat’s Last Theorem ([math]a^z+b^z=c^z[/math] has integer solution [math]a,b,c[/math] greater than 1 only for [math]z=1,2[/math] ), and that assuming what you are proving is not valid because it is circular reasoning.

 

First of all, it's not "an attempted proof of Fermat's Last Theorem" but, (with all due humility),

a "possibly valid proof of both the Beal Conjecture and Fermat's Last Theorem".

 

Now, when Craig D says,

"I pointed out in post #3 that it's necessary to assume in step (2)

that [math]z=2[/math], not [math]z=3[/math] or any greater integer value"

well.., that's where he is wrong!

It's not necessary in step (2) of my proof to make

any assumption whatsoever as to the value of [math]z[/math].

Anyone can go to step (2) in my proof and see for themselves that

step (2) allows all kinds of different values for [math]z[/math],

and that it is in steps (4) and (5) that the existence of both

logarithms and Pythagorean triples necessitate [math]z=1[/math] and [math]z=2[/math] respectively.

 

Step (2) in the proof is not an "assumption on the variables",

but rather a "statement that Pythagorean triples exist."

Basically, Craig D wants us to ignore the fact that Pythagorean triples exist!

He doesn't seem to understand that if we don't allow well known facts,

then we don't allow proofs, mathematics, or for that matter, any science whatsoever!

 

What Craig D also doesn't seem to understand is that factoring is "standard procedure",

and has absolutely nothing to do with "making an assumption",

but rather with "stating a fact"!

 

"Websters dictionary" defines the word fact as:

"Anything actually existent. Any statement strictly true; truth; reality."

and the word assumption as:

"A supposition. Something taken for granted but not necessarily true or false."

 

In fact, if only Craig D went to (Fermat's Last Theorem--from Wolfram Math World),

then he would have discovered that the very first thing we do

when faced with an equation such as:

 

[math]a^x+b^y=c^z[/math] ___________________________(11.1)

 

or

 

[math]a^n+b^n=c^n[/math] ___________________________(11.2)

 

is factor !!!!!

 

For equation (11.2), Wolfram gives:

 

[math]\left(c^\frac{n}{2}+b^\frac{n}{2}\right)\left(c^\frac{n}{2}-b^\frac{n}{2}\right)=a^n[/math]

 

and

 

[math]\left(c^\frac{n}{2}+a^\frac{n}{2}\right)\left(c^\frac{n}{2}-a^\frac{n}{2}\right)=b^n[/math]

 

from which we get:

 

[math]\left(a^\frac{n}{2}\right)^2+ \left(b^\frac{n}{2}\right)^2=\left(c^\frac{n}{2}\right)^2[/math]

 

which is, as we can plainly see, factored in exactly the same way as occurs in my proof!

 

Factoring in this manner is so common and ubiquitous,

that it is considered "par for the course", and frankly,

I'm surprised that Craig D doesn't seem to know that.

Wolfram, and virtually every book on number theory,

would not be teaching factoring if it resulted in "circular reasoning"!

 

As I said before, factoring does not, in any way, constitute an "unstated assumption",

and in order for there to actually be any kind of "circular reasoning" whatsoever,

I myself would have to make the assumption that

the exponential variables in (2) are all equal to [math]2[/math].

However, as anyone can see, I make no such assumption.

I simply allow the variables to remain variables throughout the proof,

and when it finally does occur that [math]z=1[/math] and [math]z=2[/math] are required,

that requirement is not my doing,

but an unavoidable consequence of the fact

that both the properties of logarithms and Pythagorean triples exist.

 

Originally Posted by CraigD

1) If there was known to exist a triple of integers such that [math]a^3+b^3=c^3[/math],

would the choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) be permitted?

 

2) How do you know there does not exist such a triple?

 

The first question automatically denies both Wile's proof of FLT,

and my proof of both the Beal Conjecture and FLT.

Thus, it is wildly speculative and unrealistically hypothetical.

(Kind of like asking: "If there was known to exist a human that looked exactly like a chicken,

would the choice of having it for dinner instead of having it over for dinner be permitted?")

 

My answer to that question was not "a wordy yes", but:

 

Quoting myself:

The choice of [math]3[/math] in place of (or in addition to) [math]2[/math] in (2) would then be permitted,

but we would then have to eliminate the logarithms in (4) and (5),

not to mention the rest of mathematics!

 

Since we can't eliminate the notion of logarithms from mathematics,

it must be the case that the question itself is nonsensical, which it clearly is!

 

The second question:

 

Quoting Craig D:

2) How do you know there does not exist such a triple?

 

was answered as follows:

 

Originally Posted by Don Blazys

Well, for the longest time, we didn't know!

However, if we accept the relatively recent Wiles proof of FLT,

then we can say that we "know", and pretend that we "understand",

even though the vast majority of us will never really understand it!

Thus, for those of us who are not part of the "mathematical elite",

if we simply realize that the properties of logarithms are perfectly valid,

and, if used correctly, will never ever involve us in

an inconsistency with the rest of mathematics,

then the proof at the beginning of this thread is, in my humble opinion,

more than sufficient to convince the "rest of us", that FLT is indeed true,

and that such a "triple" can't possibly exist.

 

Which Craig D took to mean:

 

Quoting craig D:

In short, Don asserts that Fermat’s Last Theorem is true because of Wiles’s proof of it.

 

As anyone can see, I asserted nothing of the sort!

I clearly said that the Wiles proof, if we accept it, (and thats a big "if"),

is far too contrived and convoluted for most people,

and that the more obvious reason we know that

such a triple can't possibly exist is because of my proof !

 

In post#4, I demonstrated that Craig D was wrong about the "indeterminate forms"

and explained that he is confusing a factorinzation for an "unstated assumption".

 

In this post, I direct him to (Fermat's Last Theorem--from Wolfram Math World).

so that he can find out for himself that

a factorization is not an "unstated assumption",

is done all the time, and does not result in "circular reasoning".

 

My proof has therefore not been logically refuted,

and I have all the faith in the world that Craig D,

along with all the other really smart math enthusiasts here at this wonderfull forum,

will, sooner rather than later,

get to the truth of this most important matter.

 

Don.

[Qfwfq]

(It's somewhat amusing to note the first four letters in "factoring" spell "fact ",

while the first three letters in "assumption "...well, you get the idea),

The words fact and factor are indeed from the same Latin root; the first means "what has been done" and the second means "one who does" (something).

 

Don you do not prove your case with this kind of conduct. :naughty:

[Don Blazys]

To: qfwfq,

 

I was only trying to "break up" a rather lenghty post with a little humor.

Clearly, I was referring to donkeys,

you know, the kind that are mentioned in the Bible.

I certainly didn't mean to offend you!

I will edit it out.

 

Don.

[Don Blazys]

To: Whoever "moved" my thread.

 

I want you to note that those who tried to refute my proof

in no way demonstrated that it contains a "fatal flaw".

 

First Craig D tried to say that it contained "indeterminate forms" that rendered it "invalid".

Then, when I demonstrated in post #4 that those

alleged "indeterminate forms" can be easily avoided,

he claimed that step (2) in my proof constitutes

an "unstated assumption" and creates a "circular argument".

Then, when I showed that factoring is allowable and in no way creates a "circular argument",

even refering him to the "Wolfram Mathworld" article on "Fermat's Last Theorem"

so that he could see for himself that factoring is quite common, my thread got "moved"!

 

Thus, to show how unfounded his "unstated assumption" claim really is,

I re-wrote my proof, using the exact same "textbook factorization" that I refered him to!

 

Now, please, follow this carefully.

 

Craig D. wrote:

 

"I pointed out in post #3 that it's necessary to assume in step (2)

that [math]z=2[/math], not [math]z=3[/math] or any greater integer value..."

 

Well, here are steps (1) and (2) in my proof:

 

[math]c^z-b^y=a^x[/math] _________________________________________________(1)

 

where factoring results in:

 

[math]\left(c^\frac{z}{2}+b^\frac{y}{2}\right)\left(c^\frac{z}{2}-b^\frac{y}{2}\right)=a^x[/math]. ___________________________________(2)

 

Notice that in step (2), we can let [math]a=8, b=8, c=4, x=3, y=3, z=5[/math],

where all of the variables are greater than [math]2[/math].

 

Thus, Craig D is mistaken, but that's okay, because even Einstein made mistakes.

 

Also note that after step (2), in steps (4) and (5),

we can't let the variables assume those same values,

because we then assume that the terms are co-prime.

 

That's as far from a "circular argument" as you can get!

 

As it stands, this proof remains a wonderful enigma,

and until it is either verified or logically refuted,

it belongs in the physics and mathematics section.

 

Don.

[Qfwfq]

Actually Don it's only in American English that the word *** means arse i. e. the posterior end, whereas referring to the notoriously stupid equine quadrupede remains a form of flaming too.

 

I lacked the time to add that this word is an old Indo-European root while assume is from the Latin ad sumere; you might well notice that the a and the s are even in separate words of the etymus. I moreover should add that, even going by your spelling-based rhetoric, I'd note that it suggests the quadrupede being he who takes the assumption as a certainty, rather than he who points this out.

 

To: Whoever "moved" my thread.

I was exercising mild humour in my manner of replying to the derogatory remark in the very start of that post, the move is however fitting with your conduct as well as manner of upholding your claim.

[Don Blazys]

To: Qfwfq,

 

The joke about what occurs when you "assume" is not even original.

I heard it a long time ago on an old T.V. show called "The Odd Couple".

Hence I thought that it was sufficiently benign and innocuous to post.

Please, let's not let our sensibilities become too refined and delicate.

Let's not forget that this is supposed to be fun.

 

Besides, it certainly wasn't "aimed" at Craig D, whose work I happen to respect.

(Indeed, it was Turtles "Strange Numbers" thread and Craig D's

"prime number algorithm" that most influenced my decision to join

this forum rather than some other science forum.)

 

While it might seem that Craig D and I (perhaps I somewhat more than he)

were rather abrupt and abrasive with each other during our debates,

I in fact appreciate that he has the courage

to challenge my viewpoint and thus put my proof through the

"trial by fire" and "acid test" that it will ultimately have to go through anyway,

in order for it to be accepted by the math community.

 

Here's how I look at it.

Both you and Craig D. are well respected mathematicians.

Thus, if my proof can withstand your best attempts to refute it,

then that will go a long way towards it's ultimate verification.

That alone should tell you that I have enormous respect

for both your and Craig D's mathematical abilities,

for if they weren't formidable, then I would be wasting my time!

The scutiny of incompetent mathematicians is absolutely worthless to me!

 

Thus, I am willing to do whatever it takes to keep my proof in the

physics and mathematics section of this fine forum.

Note that I "edited out" the "assume" joke,

the comment about Craigs "objection" being "just plain silly",

and even the "dancing smilies" that Tormod pointed out as being "inappropriate".

If there are other things in this thread that are offensive to you,

then please let me know, and I will edit them out too.

 

Please remember that I'm relatively new to posting in forums,

and "learned" in a non-science forum where rudeness was the order of the day.

As a result, I may have developed a few "bad habbits".

But (thanks to Modest) it is here that I learned to post in LaTex.

And it will be here, (I hope) that I learn to

more effectively communicate my mathematical ideas.

Think what you like about me, but please don't think me ungratefull.

 

This forum is civilized, and for me, that's like a breath of fresh air. A precious gift!

 

Please, move this thread back to the physics and mathematics section.

 

I have resolved to make it a "personal challenge"

to avoid offending anyone, ever again,

even if I myself sometimes feel offended.

 

Don.

[Don Blazys]

Quoting qfwfq:

I was exercising mild humour in my manner of replying

to the derogatory remark in the very start of that post,

 

What makes that "mild humour" genuinely funny

is that it is completely and utterly irreverent to that

"most distinguished mathematician, Professor Don Blazys".

 

I like it!

 

It's one of those rare jokes that will get even funnier as time goes on.

 

Don.