-   Miscellaneous Math (

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.
Please notify me if I made any errors.

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...

All times are UTC. The time now is 01:33.

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.