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

Awojobi · 2019-10-21, 00:31

Proof attached.

mathwiz · 2019-10-21, 00:40

Quote:

Originally Posted by Awojobi

Proof attached.

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

2M215856352p1 · 2019-10-21, 10:53

Good luck

See http://www.math.unt.edu/~mauldin/beal.html 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:

Originally Posted by Awojobi

... would not even be looked at because I am not a respected professional mathematician.

That is definitely not the reason.

Dr Sardonicus · 2019-10-21, 15:26

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.

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 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.

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 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.)

2019-10-21, 10:53	#3
2M215856352p1 May 2019 113 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

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2019-10-21, 12:17	#4
Awojobi Feb 2019 7·13 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 1000001001010₂ 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 7×13 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, 20:03	#9
Dr Sardonicus Feb 2017 Nowhere 2·3·5·139 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 91₁₀ 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 2×3×5×139 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.)