May 7th, 2016, 05:51 AM  #11  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra  Quote:
Does your paper use any mathematics that is not taught in highschool? 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.  
May 7th, 2016, 07:18 AM  #12 
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 140 Thanks: 25  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.

May 7th, 2016, 07:30 AM  #13 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
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.

May 7th, 2016, 09:02 AM  #14 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 
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 highschool mathematics, I think that you don't understand the problems. 
May 7th, 2016, 09:09 AM  #15 
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 
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. 
May 7th, 2016, 04:25 PM  #16 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
Why do you have a problem if they are irrational? You better read the whole thing before you make further comments.

May 8th, 2016, 11:06 AM  #17 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  
May 8th, 2016, 01:24 PM  #18 
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20  
May 8th, 2016, 10:38 PM  #19 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
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.

May 19th, 2016, 05:51 AM  #20 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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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