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Old 2019-10-21, 00:31   #1
Awojobi
 
Feb 2019

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

Proof attached.
Attached Files
File Type: pdf Proof of Fermat and Beal (1).pdf (323.1 KB, 85 views)
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Old 2019-10-21, 00:40   #2
mathwiz
 
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Quote:
Originally Posted by Awojobi View Post
Proof attached.
Certainly the twin primes conjecture and Riemann's hypothesis clearly follow as corollaries, yes?
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Old 2019-10-21, 10:53   #3
2M215856352p1
 
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Good luck

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

Last fiddled with by 2M215856352p1 on 2019-10-21 at 11:44
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Old 2019-10-21, 12:17   #4
Awojobi
 
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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.
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Old 2019-10-21, 13:15   #5
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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.
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Old 2019-10-21, 14:38   #6
Awojobi
 
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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.
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Old 2019-10-21, 14:55   #7
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Quote:
Originally Posted by Awojobi View Post
... would not even be looked at because I am not a respected professional mathematician.
That is definitely not the reason.
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Old 2019-10-21, 15:26   #8
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Quote:
Originally Posted by Awojobi View Post
It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red.




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Old 2019-10-21, 20:03   #9
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Your first argument, beginning "Let it be initially assumed that A and B have a highest common factor = 1..." does not 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.
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Old 2019-10-22, 11:06   #10
Awojobi
 
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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.
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Old 2019-10-22, 11:58   #11
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I was not trying to be funny. Your argument that begins, "Let it be initially assumed that A and B have a highest common factor = 1" does not use the hypothesis that the exponents x and y are greater than 2. Anywhere.

And, as I have already shown, that means your argument leads to a demonstrably false conclusion.

So, that flaw kills your proof so dead, you'll need two graves to bury it. (OK, that 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.)
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