mersenneforum.org "PROOF" OF BEAL'S CONJECTURE & FERMAT'S LAST THEOREM
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2019-10-21, 00:31   #1
Awojobi

Feb 2019

7·13 Posts
"PROOF" OF BEAL'S CONJECTURE & FERMAT'S LAST THEOREM

Proof attached.
Attached Files
 Proof of Fermat and Beal (1).pdf (323.1 KB, 105 views)

2019-10-21, 00:40   #2
mathwiz

Mar 2019

11·13 Posts

Quote:
 Originally Posted by Awojobi Proof attached.
Certainly the twin primes conjecture and Riemann's hypothesis clearly follow as corollaries, yes?

 2019-10-21, 10:53 #3 2M215856352p1   May 2019 7116 Posts 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
 2019-10-21, 12:17 #4 Awojobi   Feb 2019 5B16 Posts 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.
 2019-10-21, 13:15 #5 Dr Sardonicus     Feb 2017 Nowhere 25·131 Posts 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.
 2019-10-21, 14:38 #6 Awojobi   Feb 2019 1338 Posts 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.
2019-10-21, 14:55   #7
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

100100001100012 Posts

Quote:
 Originally Posted by Awojobi ... would not even be looked at because I am not a respected professional mathematician.
That is definitely not the reason.

2019-10-21, 15:26   #8
Dr Sardonicus

Feb 2017
Nowhere

25×131 Posts

Quote:
 Originally Posted by Awojobi It is not an assumption. It is justified because the 2 sides of the equation have the same corresponding expressions highlighted in red.

 2019-10-21, 20:03 #9 Dr Sardonicus     Feb 2017 Nowhere 101408 Posts 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.
 2019-10-22, 11:06 #10 Awojobi   Feb 2019 7×13 Posts 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.
 2019-10-22, 11:58 #11 Dr Sardonicus     Feb 2017 Nowhere 106016 Posts 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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