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May 7th, 2016, 04:51 AM   #11
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Quote:
Originally Posted by MrAwojobi View Post
The so called maths professionals have the same attitude i.e. they are in denial that the solution to both similar problems is just as simple as the quote of the problems themselves.
That's right. A mathematical journal, whose livelihood depends on the interest and quality of the articles it publishes, doesn't want to publish a correct proof of two of the biggest problems in mathematics.

Does your paper use any mathematics that is not taught in high-school? I very much doubt it (because of the nature of such attempted proofs generally). It's inconceivable that a journal of such prestige could fail to understand your proof, and if it were correct, it it inconceivable that they wouldn't print it.
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May 7th, 2016, 06:18 AM   #12
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Quote:
Originally Posted by MrAwojobi View Post
Why don"t you simply go through my work yourself and then acknowledge that this is the
proof.
Do you agree to acknowledge that it isn't a proof, if we find a flaw in it? History seems to indicate that you won't, so why should we bother? It has to work both ways.
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May 7th, 2016, 06:30 AM   #13
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You are demonstrating the same attitude of being in denial. One would doubt if this is a maths forum.Maths is supposed to be discussd here not personalities. Read my proof and make comments that are based on mathematics.
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May 7th, 2016, 08:02 AM   #14
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I would characterise my attitude as disinterested or impartial. I have no reason to deny your claims if they are correct, and neither has the magazine.

As with most people who claim to have solved these problems in a couple of pages of high-school mathematics, I think that you don't understand the problems.
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May 7th, 2016, 08:09 AM   #15
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Very interesting. It starts with
A^x+B^y=C^z
Then appears
A^x+B^y=M^n i.e. C^z=M^n
and M is assumed to be integer as well as A^x/n and B^y/n.
But if n does not divide x, y, z all these numbers are irrational.
There's no reason to read the rest.
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May 7th, 2016, 03:25 PM   #16
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Why do you have a problem if they are irrational? You better read the whole thing before you make further comments.
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May 8th, 2016, 10:06 AM   #17
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Quote:
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Why do you have a problem if they are irrational? You better read the whole thing before you make further comments.
If you think being irrational is NOT a problem, you had better learn what "Fermat's Last Theorem" and "Beal's Conjecture" really say!
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May 8th, 2016, 12:24 PM   #18
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Quote:
Originally Posted by MrAwojobi View Post
Why do you have a problem if they are irrational? You better read the whole thing before you make further comments.
It's assumed in your paper that they are rational. Did you read "the whole thing" yourself?
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May 8th, 2016, 09:38 PM   #19
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Can both of you not see that for the case of FLT the terms are integers to the power of n. All I have done is reduce the problem to something that looks like an identity but shown not to be simply because of the impossibillity of equating coefficients.
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May 19th, 2016, 04:51 AM   #20
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Yes, integers required. But your reference to "irrational numbers" immediately assumes that cannot true- you are assuming from the start that FLT is true.
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