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-   -   "PROOF" OF BEAL'S CONJECTURE & FERMAT'S LAST THEOREM (https://www.mersenneforum.org/showthread.php?t=24865)

 Awojobi 2019-10-21 00:31

"PROOF" OF BEAL'S CONJECTURE & FERMAT'S LAST THEOREM

1 Attachment(s)
Proof attached.

 mathwiz 2019-10-21 00:40

[QUOTE=Awojobi;528452]Proof attached.[/QUOTE]

Certainly the twin primes conjecture and Riemann's hypothesis clearly follow as corollaries, yes?

 2M215856352p1 2019-10-21 10:53

Good luck

See [url]http://www.math.unt.edu/~mauldin/beal.html[/url] for how and where to submit the proof and claim the prize

 Awojobi 2019-10-21 12:17

My proof, just like some others that have been published in journals, would not even be looked at because I am not a respected professional mathematician. The purpose of me posting here is for a good critique, if any.

 Dr Sardonicus 2019-10-21 13:15

Beginning on the fourth line from the bottom of Page 1, you assume that the numerical equality of two expressions implies equality of corresponding coefficients in the algebraic formulations.

This assumption is not justified.

 Awojobi 2019-10-21 14:38

It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red. It therefore means equating corresponding coefficients is justified. Of course when this is done, contradictions galore begin to arise which shows that the original assumption of an equation is contradicted, given the conditions stated in the proof. Herein lies the proof.

 Batalov 2019-10-21 14:55

[QUOTE=Awojobi;528473]... would not even be looked at because I am not a respected professional mathematician. [/QUOTE]
That is definitely not the reason.

 Dr Sardonicus 2019-10-21 15:26

[quote=Awojobi;528484]It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red.[/quote]

:orly owl:

:missingteeth:

 Dr Sardonicus 2019-10-21 20:03

Your first argument, beginning "Let it be initially assumed that A and B have a highest common factor = 1..." does [i]not[/i] use the condition that x > 2 and y > 2. Therefore, if your argument were valid, it would follow that A^2 + B^2 = C^n had no solutions with n > 2 odd.

However, 2^2 + 11^2 = 5^3.

This is a counterexample to your purported proof, but not to Beal's conjecture.

 Awojobi 2019-10-22 11:06

You are trying to be funny because you know that x and y are greater than 2. It is stated in the statement of Beal's conjecture. Look for better flaws in my proof, if any.

 Dr Sardonicus 2019-10-22 11:58

I was [i]not[/i] trying to be funny. Your argument that begins, "Let it be initially assumed that A and B have a highest common factor = 1" does [i]not[/i] use the hypothesis that the exponents x and y are greater than 2. Anywhere.

So, that flaw kills your proof so dead, you'll need two graves to bury it. (OK, [i]that[/i] was me trying to be funny. I find it very sad that you seem incapable of understanding the very basic point I'm making, but I'd rather laugh than cry.)

 2M215856352p1 2019-10-23 00:18

[QUOTE=Awojobi;528484]It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red. It therefore means equating corresponding coefficients is justified. Of course when this is done, contradictions galore begin to arise which shows that the original assumption of an equation is contradicted, given the conditions stated in the proof. Herein lies the proof.[/QUOTE]

Ultimately, this is a wrong justification. This is a rather glaring mistake if you are observant enough.

The expressions you were talking about were not polynomials in d and e, but you defined d and e to be integers, not polynomial variables, hence you cannot just compare coefficients.

Not true in general does not mean there does not exist some h=i such that the expression somewhat holds by fluke, especially in the case where n is odd when the terms have alternating sign and they somehow cancel each other. Just one counterexample and the conjecture is false.

 retina 2019-10-23 02:14

[QUOTE=Awojobi;528473]My proof ... would not even be looked at because I am not a respected professional mathematician.[/QUOTE]You have no proof, therefore you get no looks.[QUOTE=Awojobi;528473]The purpose of me posting here is for a good critique ...[/QUOTE]If you were more willing to accept critique instead of shouting people down then people might be more willing to post some.

 Dr Sardonicus 2019-10-23 11:35

[quote=2M215856352p1;528638][quote=Awojobi;528484]It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red.[/quote]Ultimately, this is a wrong justification. This is a rather glaring mistake if you are observant enough.[/quote]Yes, indeed! Everyone knows that, in order for a proof by highlighting to be valid, highlights should be in [i][color=yellow][size=4]yellow[/size][/color]![/i]
:missingteeth:

 Awojobi 2019-10-23 13:06

I don't understand what you mean when you talk about d and e not being polynomial variables and therefore I cannot compare coefficients. Comparison of coefficients is justified in order to establish the equality of LHS and RHS of the equation since the expressions highlighted in red are the same for RHS and LHS. You also talk about counterexamples. Counterexamples of what? Whatever counterexamples you are talking about, produce them.

 LaurV 2019-10-23 13:08

[QUOTE=Dr Sardonicus;528677]:missingteeth:[/QUOTE]
:missingteeth:

 10metreh 2019-10-23 14:29

[QUOTE=Awojobi;528682]I don't understand what you mean when you talk about d and e not being polynomial variables and therefore I cannot compare coefficients. Comparison of coefficients is justified in order to establish the equality of LHS and RHS of the equation since the expressions highlighted in red are the same for RHS and LHS.[/QUOTE]

Consider the equation 3[COLOR="Red"]d[/COLOR]+4[COLOR="red"]e[/COLOR] = 6[COLOR="red"]d[/COLOR]+2[COLOR="red"]e[/COLOR].
This is true for the specific values d = 2 and e = 3, for example. This does not allow us to conclude that 3 = 6 and 4 = 2.

(Also from what I can tell the "proof" never used the fact that A, B and C are all integers?)

 2M215856352p1 2019-10-24 06:32

[QUOTE=Awojobi;528682]I don't understand what you mean when you talk about d and e not being polynomial variables and therefore I cannot compare coefficients. Comparison of coefficients is justified in order to establish the equality of LHS and RHS of the equation since the expressions highlighted in red are the same for RHS and LHS. You also talk about counterexamples. Counterexamples of what? Whatever counterexamples you are talking about, produce them.[/QUOTE]

Let n=3. It follows that,
A[SUP]x[/SUP] + B[SUP]y[/SUP] = (A[SUP]x/3[/SUP])[SUP]3[/SUP] + (B[SUP]y/3[/SUP])[SUP]3[/SUP] = [(A[SUP]x/3[/SUP]) + (B[SUP]y/3[/SUP])][A[SUP]x/3[/SUP])[SUP]2[/SUP] - (A[SUP]x/3[/SUP])(B[SUP]y/3[/SUP]) + (B[SUP]y/3[/SUP])[SUP]2[/SUP]] = M[SUP]3[/SUP]

It is defined that M = d - e, A[SUP]x/3[/SUP] = hd, B[SUP]y/3[/SUP] = ie, where d and e are positive integers and h and i are positive real numbers, if I have not mistaken.

By division by M from both sides, we end up getting
[(hd + ie)/(d - e)]h[SUP]2[/SUP] d[SUP]2[/SUP] - [(hd + ie)/(d - e)]hi de + [(hd + ie)/(d - e)]i[SUP]2[/SUP] e[SUP]2[/SUP]
= d[SUP]2[/SUP] - 2de + e[SUP]2[/SUP]

Comparing coefficients of d[SUP]2[/SUP]: [(hd + ie)/(d - e)]h[SUP]2[/SUP] = 1
Comparing coefficients of de: -[(hd + ie)/(d - e)]hi = -2
Comparing coefficients of e[SUP]2[/SUP]: [(hd + ie)/(d - e)]i[SUP]2[/SUP] = 1

Hence, indeed there is a contradiction by comparing coefficients. However, weird examples like 2^2+11^2=5^3 can appear because it does not need to be the case where each term is equal in value for a specific value of d and e.

A counterexample to Beal’s conjecture is very hard to find. Till now, we still have not found any yet.

Awojobi's variable names spell hide, he seems to be hiding something, isn't he? (joking)

 axn 2019-10-24 08:07

[QUOTE=2M215856352p1;528753]A counterexample to Beal’s conjecture is very hard to find. [/QUOTE]
Understatement of the century. If we ever do find a counterexample, the conjecture will be falsified. So asking for a counterexample is to falsify the conjecture itself and not just the proof.

However, while the proof itself say that A,B,M, x,y,d,e are integers and h&i are rationals, none of the algebraic manipulations require it to be so (i.e. uses these properties). So we can pick arbitrary values for these -- specifically, we can set M= (A^x+B^y)^(1/3) and observe the behavior of the coefficients. OP will handwave these away, but that's SOP for him.

[code]
A=5
B=7
x=3
y=3

M=(A^x+B^y)^(1/3) /// An M that works

e=1; /// arbitrary
d=e+M

h=A/d
i=B/e

c1=(A+B)/M*h^2 ///x/3 = y/3 = 1
c2=-(A+B)/M*h*i
c3=(A+B)/M*i^2

d^2-2*d*e+e^2
c1*d^2+c2*d*e+c3*e^2

/// Look at that! Eventhough coefficients are different, end value is same!
[/code]

 Awojobi 2019-10-24 12:06

So 2M215856352p1 we are in agreement. The example you gave doesn't count be cause 2 of the powers are 2 and Beal's conjecture clearly states that the powers should be greater than 2. The other poster coming up with 3d+4e = 6d+2e to try and debunk my rationale for equating coefficients should note that the equation simplifies to 3d = 2e. There are no coefficients to compare now. Also, it is clearly stated in the Beal equation that A, B and C are all integers.

 Dr Sardonicus 2019-10-24 13:13

[quote=Awojobi;528758]The example you gave doesn't count be cause 2 of the powers are 2 and Beal's conjecture clearly states that the powers should be greater than 2.[/quote]Hypotheses in Beale's conjecture don't count if your argument doesn't use them.

 2M215856352p1 2019-10-24 13:17

[QUOTE=Awojobi;528758]So 2M215856352p1 we are in agreement. The example you gave doesn't count be cause 2 of the powers are 2 and Beal's conjecture clearly states that the powers should be greater than 2. The other poster coming up with 3d+4e = 6d+2e to try and debunk my rationale for equating coefficients should note that the equation simplifies to 3d = 2e. There are no coefficients to compare now. Also, it is clearly stated in the Beal equation that A, B and C are all integers.[/QUOTE]

Not exactly, your proof is incomplete and I can sense some ego in your dismissal of the poster who sent 3d+4e=6d+2e. You can compare coefficients in 3d=2e, it’s just meaningless, 3=0 and 0=2.

On my part, it was a completely failed attempt to come up with a counterexample. I actually disagree with you.

I still have one question to ask: why is it necessary that x, y, z, M are positive integers? It is stated in the conjecture but you have yet to use those properties in your proof. axn found a counterexample in your logic of the proof but not the conjecture. Hence I need you to explain why the logic in your proof works when x, y, z, M are positive integers, x,y,z>2 without stating that it is in the conjecture. Otherwise, that would be a circular argument.

 Awojobi 2019-10-24 13:41

No one can come up with a counterexample to disprove the truth of Beal's conjecture because I have proved it. The poster you referred to never came up with any counterexample as I have shown in a previous post what his original equation simplifies to.
I don't understand what you mean when you say I haven't used in my proof the fact that the variables you mentioned are positive integers. Beal's conjecture says they are positive integers and so what is the problem? If you look at my proof of FLT, I believe I have made a statement as to why if 2 of the powers are 2 then my whole argument brakes down i.e. no coefficients to compare just like there were no coefficients to compare in the other posters failed equation which I simplified.

 2M215856352p1 2019-10-24 15:05

Looks like Awojobi is missing the point, I shall give a quadratic counterexample.

Let d=5 and e=2.
3d+4e^2+0e = 3.8d+0e^2+6e happens to be the same for this particular set of d and e, but it is not the same for all d and e. Yet, by comparing coefficients of d, you can’t say that 3=3.8. By comparing coefficients of e^2, you get 4=0. That make s no sense at all. And you find a way to hand wave your way through because you believe that you are always right.

You compared coefficients to show that two polynomial expressions which happen to be equal for some particular values of h,i,d,e cannot be equal in general (up to there still ok), and hence get a contradiction that they cannot be equal. Wrong. Not equal in general does not mean not equal for any set of possible h,i,d,e.

It’s the wrong concept to say that if some polynomial power does not appear on one side of the equation, you cannot compare coefficients. You can compare coefficients, just that you are comparing it with 0.

Conclusion: Awojobi probably needs to relearn secondary school/high school math concepts

 LaurV 2019-10-24 15:25

:raman:

 Awojobi 2019-10-24 16:14

Another wrong example given, yet again. Your coefficients are actual numbers. My coefficients are variables. I have equated corresponding coefficients (which, I emphasize again, are variables) and loads of contradictory values for each one of the variable begin to arise.

 axn 2019-10-24 16:20

[QUOTE=Awojobi;528799]Another wrong example given, yet again. Your coefficients are actual numbers. My coefficients are variables. I have equated corresponding coefficients (which, I emphasize again, are variables) and loads of contradictory values for each one of the variable begin to arise.[/QUOTE]

You're absolutely correct. You may now move onto proving Twin Prime Conjecture and Riemann hypothesis.

 Dr Sardonicus 2019-10-24 16:35

If you simply use the author's definitions, the invalidity of the "equating coefficients" argument becomes manifest. The hypotheses are,

n = odd positive integer, M = positive integer, d, e = positive rational numbers such that d - e = M.

Then

$$A^{\frac{x}{n}}\; = \; hd\text{, }B^{\frac{y}{n}} \; = \; ie\text{.}$$

OK, so that means we can solve for h and i in terms of the other variables:

$$h\;=\;\frac{A^{\frac{x}{n}}}{d}\text{, }i\;=\;\frac{B^{\frac{y}{n}}}{e}$$

Substituting these expressions into the author's equation will give a valid identity -- regardless of whether the expressions for h and i are equal.

 Awojobi 2019-10-24 16:50

I don't get the point the last poster is trying to make. What invadility of equating coefficients are you talking about. Can you not see that when corresponding coefficients are equated, conflicting values of each variable begin to arise?

 Batalov 2019-10-24 17:09

[COLOR="Red"]MOD: This thread is becoming circular and of a troll-feeding kind.
After a couple more pointless circles we will be compelled to lock it.[/COLOR]

 Uncwilly 2019-10-24 20:43

[QUOTE=Batalov;528814][COLOR="Red"]After a couple more pointless circles we will be compelled to lock it.[/COLOR][/QUOTE]
Pointless circles -> :spinner:

 2M215856352p1 2019-10-25 00:50

[QUOTE=Awojobi;528808]I don't get the point the last poster is trying to make. What invadility of equating coefficients are you talking about. Can you not see that when corresponding coefficients are equated, conflicting values of each variable begin to arise?[/QUOTE]

This one explains why your equation substitution is correct, in support of your proof, but at the same time, also explaining why [B]comparing coefficients will lead to the wrong conclusion that that could not be the case, hence in the end you ended up contradicting not anybody but yourself.[/B]

I believe that the purpose of you sending random crank proofs is to convince mathematical fools in this forum that you are right, instead of a good critique. When you said good critique, I think you actually meant a confirmation from everybody in this forum that you are right.

People have been pointing out all the errors but you have been rudely dismissing them as if you are the most powerful, with your own ridiculous theories, which in turn were already pointed out to be wrong, which makes you look even more like a mathematical fool. This is against the spirit of mathematical research.

If Awojobi continues that way, he will forever be a crank, a troll and a mathematical fool.

Sorry for the harshness, but I think I have lost all my patience on Awojobi. [B]Time to stop feeding the troll.[/B]

To help you, if you don’t understand a point, please do not reiterate the main idea of your proof because people will see that you are insistent and not willing to learn new things.

 chalsall 2019-10-25 01:19

[QUOTE=Uncwilly;528837]Pointless circles -> :spinner:[/QUOTE]

It always amuses me when I see something like that.

Developer: "Let's keep the human engaged by showing them this animated GIF which we've already downloaded. Absolutely no correlation with what is actually going on, but most won't understand that...

It's like when VoIP finally got good, and we had to start introducing "comfort noise" into the streams... Sigh...

 2M215856352p1 2019-10-25 05:04

Proof that Awojobi’s proof is garbage

Suppose, on the contrary, that Awojobi managed to prove Beal’s conjecture.

Now, let’s generalise Beal’s conjecture!

Let A,B,M be positive integers and n be an integer at least three. Let x,y be integers > 0. Prove or disprove that if A^x + B^y = M^n, then A,B and M must have a highest common factor greater than 1.

**Shamelessly copies Awojobi’s ‘proof’ of Beal’s conjecture and pastes it here, since the logic works perfectly.** Since Beal’s conjecture is a subset of this conjecture, Beal’s conjecture must be true.

However, one can easily contrive an example to disprove that. 2^2+11^2=5^3. Contradiction. Therefore, Awojobi’s proof is wrong.

This is my question:

[B]Without reiterating the steps of your proof, stating that it is due to the conditions of the conjecture or any form of stubborn insistence that you are correct[/B], please explain why your proof does not work for x < 2 or y < 2.

If you violate the part in bold, I will have to ignore you for one week for this thread to protect our own sanity. I’m sorry for being like a teacher who tends to nag. Only trolls will do the part in bold and I cannot convince any of the people in this category. The moderators have described you as a troll and all others in this forum as troll-feeders.

 2M215856352p1 2019-10-25 11:16

Oops, for the question, I actually meant x<=2 or y <=2 rather than x < 2 or y < 2

 Dr Sardonicus 2019-10-25 11:40

I'm done with this thread. It was good practice in assessing argumentation.

However, the range was limited. A step reminiscent of an old Sidney Harris cartoon ("Then a miracle occurs"). A failure to use hypotheses.

My posts to this thread were predicated on the hypothesis that the OP was interested in learning.

The OP's responses have convinced me that this hypothesis is wrong, and that the OP is merely trolling. I am therefore also done not only with this thread, but with this poster.

The OP can expect no further responses from me. Ever.

 Awojobi 2019-10-25 12:34

2M215856352p1 the points you raise are pointless. If you cannot see that if x and or y and or z are 2 or less, then there are no coefficients to compare, sorry I can't help you any further. If you can't see that if the variables have a highest common factor, the form of the equations cannot be altered, then I can't help you. The only time the form of the equations
can be altered is if the variables have a highest common factor that is greater than 1. In this situation the equation morphs into a different form such that equating coefficients can no longer happen i.e. Beam's equation can only happen if the highest common factor is greater than 1.

 Awojobi 2019-10-25 12:36

.

 2M215856352p1 2019-10-25 12:42

[QUOTE=Awojobi;528890]2M215856352p1 the points you raise are pointless. If you cannot see that if x and or y and or z are 2 or less, then there are no coefficients to compare, sorry I can't help you any further. [/QUOTE]

Please read your proof carefully. I think you misinterpreted your own proof, which is the worst thing I could have imagined in this thread. x and y can be equal to two for the logic in your proof to work, only when z=2 then there are still coefficients to compare, just no contradiction. It’s the value of z that dictates the polynomial expansion. When z=2, I understand where you are coming from.

 Awojobi 2019-10-25 12:42

Highest common factor of 1, I meant to say.

 Awojobi 2019-10-25 12:48

You are not even being coherent enough to say exactly what problems you have with my proof. You have not given any good mathematical reason why it doesn't make very good sense for coefficients to be compared. You're just being dismissive because you are unable to disprove all my proofs of maths conjectures.

 mathwiz 2019-10-25 12:50

Okay, okay, you've convinced us! Clearly the proof is airtight and it's time to close this thread and move on to other things....

 retina 2019-10-25 12:50

Hehe, none so blind as those that refuse to see.

 2M215856352p1 2019-10-25 12:51

[QUOTE=2M215856352p1;528753]Let n=3. It follows that,
A[SUP]x[/SUP] + B[SUP]y[/SUP] = (A[SUP]x/3[/SUP])[SUP]3[/SUP] + (B[SUP]y/3[/SUP])[SUP]3[/SUP] = [(A[SUP]x/3[/SUP]) + (B[SUP]y/3[/SUP])][A[SUP]x/3[/SUP])[SUP]2[/SUP] - (A[SUP]x/3[/SUP])(B[SUP]y/3[/SUP]) + (B[SUP]y/3[/SUP])[SUP]2[/SUP]] = M[SUP]3[/SUP]

It is defined that M = d - e, A[SUP]x/3[/SUP] = hd, B[SUP]y/3[/SUP] = ie, where d and e are positive integers and h and i are positive real numbers, if I have not mistaken.

By division by M from both sides, we end up getting
[(hd + ie)/(d - e)]h[SUP]2[/SUP] d[SUP]2[/SUP] - [(hd + ie)/(d - e)]hi de + [(hd + ie)/(d - e)]i[SUP]2[/SUP] e[SUP]2[/SUP]
= d[SUP]2[/SUP] - 2de + e[SUP]2[/SUP]

Comparing coefficients of d[SUP]2[/SUP]: [(hd + ie)/(d - e)]h[SUP]2[/SUP] = 1
Comparing coefficients of de: -[(hd + ie)/(d - e)]hi = -2
Comparing coefficients of e[SUP]2[/SUP]: [(hd + ie)/(d - e)]i[SUP]2[/SUP] = 1

Hence, indeed there is a contradiction by comparing coefficients. However, weird examples like 2^2+11^2=5^3 can appear because it does not need to be the case where each term is equal in value for a specific value of d and e.

A counterexample to Beal’s conjecture is very hard to find. Till now, we still have not found any yet.

Awojobi's variable names spell hide, he seems to be hiding something, isn't he? (joking)[/QUOTE]

If you let x=y=2, there is no problem with the logic above. I do not see any issues comparing coefficients. Do you need me to replace all the x and y with 2?

EDIT: This thread is finally going somewhere...

 Awojobi 2019-10-25 13:05

So let me get you right. Are you saying you want me to explain why your so called weird example doesn't follow Beam's conjecture?

 Awojobi 2019-10-25 13:06

So let me get you right. Are you saying you want me to explain why your so called weird example doesn't follow Beal's conjecture?

 Awojobi 2019-10-25 13:19

The explanation to your so called weird example is simply that it cannot be factorised I.e. it cannot be written with brackets that have integer values.

 2M215856352p1 2019-10-25 13:54

[QUOTE=Awojobi;528901]The explanation to your so called weird example is simply that it cannot be factorised I.e. it cannot be written with brackets that have integer values.[/QUOTE]

Let x=y=5 and z=3. If that is the case, then your proof really boils down to nothing since 5/3 is not integer.

Looks like the thread could be locked to protect our sanity.

 axn 2019-10-25 13:58

[QUOTE=mathwiz;528895]it's time to close this thread and move on to other things....[/QUOTE]
Agreed

 Uncwilly 2019-10-25 14:01

[QUOTE=2M215856352p1;528905]Looks like the thread could be locked to protect our sanity.[/QUOTE]I was in the process of doing that when your post came in.

 Batalov 2019-11-02 16:50

Randall explains it all...

A quick bit of staircase wit - it explains this whole thread in one picture:
[url]https://xkcd.com/2217/[/url]

[SPOILER]these days I watch xkcd in large batches ... and find something that could have been useful somethime ago[/SPOILER]

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