[QUOTE=rogue;261005]Your hand waving continues when you state that your proof "predicts" other statements without any evidence. It almost appears that you are trying to "induct" other "truths" from your proof for equations with more than three terms. Others have already shown the flaws of that logic. You tend to then "amend" your proof so that "these other conditions must also apply" without stating anything about how you arrived at such conditions.[/QUOTE]
What's more, when Don does suggest these "other conditions," they actually contradict other elements in his "proof." I am beginning to believe that Don doesn't actually understand what he thinks his "proof" does. He somehow accepts its "truth" on faith, so any criticism of it must be wrong, and any argument supporting it must be right even though he can't say why.

For example, "In my proof, the term under the radical applies to any sum or difference of two terms only." No, it applies to any equation where he can isolate one term of the form C^Z on one side, as a function of the others. "My proof predicts that with four terms, at least two of them will always have a common factor if w,x,y,z > 2." This is the only correct thing he has said. His "proof," if correct, does predict this. Too bad it is wrong, as proven by example. For third powers and now, thanks to Elkies via Silverman, fourth powers. "For one thing, a,b,c,d must have no common factor whatsoever. In other words, a,b,c,d must be pairwise co-prime, as are a,b,c in Beal's Conjecture." No, he specifically used the fact that the only factor that is common [i]all]/i] of the terms is 1=(T/T). Nowhere does he isolate just two. "What I meant to write was w,x,y,z > 3." This, of course, requires that Z must be "allowed" to be 1, 2, [b]or 3[/b], and "disallowed" only when Z>3. Yet he steadfastly insists that 3 is disallowed. But I suppose he will know raise that to "What I meant to write was w,x,y,z > 4." Guess what - 5 works, too.

And no, it is not a matter of not having time to respond to every comment. While I agree that can be a valid argument when faced with similar comments, it isn’t when they are dissimilar as in this discussion. He simply does not understand enough of the math to fabricate even his usual incorrect arguments, so he ignores them.

So, Don, reply to this one point. Use real, supportable answers, not "because it must be so!"[indent]Why does choosing M=1 and M=2 in posts #260, #286, and #315 lead to "allowing" Z=1 and Z=2, but any other arbitrary M does not lead to "allowing" Z=M and "disallowing" Z=1 and Z=2?[/indent]

[QUOTE=R.D. Silverman;261010]I missed this post earlier.......

Solutions are already known. In fact, infinitely many are known. Noam
Elkies found the first one. Roger Frye found the smallest positive solution.
[findit it was a non-trivial search]. Solutions are known to be dense in the
rationals.[/QUOTE]

Interesting. I did another one of those 10 minute hacks to look for one but it was unsuccessful. I thought the problem might be related to integer points on hyperelliptic curves but I don't really know anything about them and so can't use their theory in the search. Do you happen to have a pointer or searchable buzzword for these results?

[QUOTE=R.D. Silverman;261010]I missed this post earlier.......

Solutions are already known. In fact, infinitely many are known. Noam
Elkies found the first one. Roger Frye found the smallest positive solution.
[findit it was a non-trivial search]. Solutions are known to be dense in the
rationals.[/QUOTE]You also missed the subsequent weaseling out of the claim. The new claim is that the integers must be pair-wise coprime.

I had a quick look around and couldn't find any which meet the additional constraint. That's not to say that none have been published, only that I couldn't find them in the limited amount of effort I put into the task.

Paul

[QUOTE=xilman;261019]You also missed the subsequent weaseling out of the claim. The new claim is that the integers must be pair-wise coprime.

I had a quick look around and couldn't find any which meet the additional constraint. That's not to say that none have been published, only that I couldn't find them in the limited amount of effort I put into the task.

Paul[/QUOTE]

I suspect that the existence of pairwise co-prime solutions can be proven
from the fact that the set of fractions formed from the solutions is dense
in the rationals.

[QUOTE=akruppa;261018]Interesting. I did another one of those 10 minute hacks to look for one but it was unsuccessful. I thought the problem might be related to integer points on hyperelliptic curves but I don't really know anything about them and so can't use their theory in the search. Do you happen to have a pointer or searchable buzzword for these results?[/QUOTE]

Search on the two names that I gave. The solutions are indeed derived
from integer points on a (pencil of) elliptic [not hyperelliptic] curves.

[QUOTE=xilman;261019]You also missed the subsequent weaseling out of the claim. The new claim is that the integers must be pair-wise coprime.[/quote]
But it is a meaningless claim unless he can point to a part of his proof that uses it that way. There are none. He uses it to assert that teh only factor common to all of the terms is 1.

[QUOTE=Condor;261028]But it is a meaningless claim unless he can point to a part of his proof that uses it that way. There are none. He uses it to assert that teh only factor common to all of the terms is 1.[/QUOTE]Sure. Is English your native language and are you cognizant of the meaning of the colloquial verb "to weasel" or of the noun phrase "weasel words"?


Paul

[QUOTE=Don Blazys;260978]Like I said so many times before, if there was even one "fatal flaw" in my proof, then this thread would have ended a long, [B][I]long[/I][/B] time ago![/QUOTE]

This is the new math -- proof by number of Internet posts!!! I didn't learn this method in college, could you enlighten me as to how it works? Is it the total number of posts that count, or is it posts per day?? How many posts do I need to establish a proof??

I am eager to use this wonderful new technique to prove Goldbach's conjecture [URL="http://mersenneforum.org/showthread.php?t=7592"]http://mersenneforum.org/showthread.php?t=7592[/URL], the existence of free energy [URL="http://mersenneforum.org/showthread.php?t=6236"]http://mersenneforum.org/showthread.php?t=6236[/URL], the end of the world [URL="http://mersenneforum.org/showthread.php?t=15582"]http://mersenneforum.org/showthread.php?t=15582[/URL], and so much more!!

Thank you Don for allowing me to contribute to the proofiness of your claim by adding one more post. I trust I can count on you to increase the proofiness of any claims I make in the future!

Fourth powers are 0,1 (mod 5). If d^4 == 1 (mod 5), the LHS must consist of two summands that are 0 (mod 5) and one that is 1 (mod 5). If d^4 == 0 (mod 5), the only way that the LHS sums to 0 (mod 5) is by all three summands being 0 (mod 5). So for w=x=y=z=4, I think there will always be a pair of bases with common factor 5. With any exponent >4, there might be coprime solutions.

I think it's time to lock the thread or leave it dead as happened in the following forums:

[url]http://www.physicsforums.com/showthread.php?t=301139[/url]
[url]http://www.sciencechatforum.com/viewtopic.php?f=19&t=11838&start=30[/url]
[url]http://www.marilynvossavant.com/forum/viewtopic.php?p=14565&highlight=#14565[/url]
[url]http://www.scienceforums.net/topic/32453-the-most-important-equation-ever/[/url]
[url]http://www.mymathforum.com/viewtopic.php?f=40&t=7283[/url]
[url]http://www.sciforums.com/showthread.php?t=103447[/url]
[url]https://nrich.maths.org/discus/messages/147612/149003.html?1273932598[/url]

and so on...

Some of us kinda enjoy this. Admittedly it's on the same level as laughing at the handicapped kid during recess, but I'm not choosy about entertainment.