"PROOF" OF BEAL'S CONJECTURE & FERMAT'S LAST THEOREM
1 Attachment(s)
Proof attached.

[QUOTE=Awojobi;528452]Proof attached.[/QUOTE]
Certainly the twin primes conjecture and Riemann's hypothesis clearly follow as corollaries, yes? 
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 
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.

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

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.
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, [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.) 
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