Recent content by vantheman

I am assuming the stronger position - n does not divide x, y or z and x, y and z have no common factors. Since n does not divide y , n does not divide h, since y == h mod n. Therefore h cannot have any n factors.

Most with any knowledge of Number Theory are aware that many hundreds of thousands of hours have been spent reviewing flawed "proofs" of Fermat's Last Theorem (FLT). It is understandable that a serious mathematician would not spend more than a few minutes looking at a possible "proof". Wiles'...

Thank you. It's very much appreciated.

Sorry for the errors. The correct web address is [PLAIN]www.mathpages.com/home/kmath367.htm[/URL] The equal sign in the formula should have been "+" not "=". I neglected to hit the shift key. I think p-3 is correct. When you substact x^p and y^p from the expansion, the results have a...

Reference: [PLAIN]www.mathpages.com/home367.htm[/URL] On page 2 of reference the formula is given (x+y)^p - x^p - y^p = pxy(x=y)Q(x,y) where Q(x,y) is a homogenous integer function of degree p-3. If we insert a number of different value of p into the equation, it appears that Q(x,y) =...

To continue, if n=3 then f(z,k) = (z-k) = y. We have established that f(z,k) must be divisible by n, but we have also established that y is not divisible by n - thus we have a contradiction. Is this not a proof of FLT for n=3?

I am not there yet, but let me review where I am and where I'm going. So far I think my paper shows that, if an integer solution to FLT exists, then the nth roots of (z-y), (z-x) and (x+y) are all integers; that neither x, y or z have a factor of n and therefore the classical 'Case 2' of FLT is...

I agree. You are right. I had two approaches to a solution to the problem. I chose one that had a fatal flaw. I am re-writing from Para. 2.4 on. Expect to post it next week. I believe the intermediate proof that is in my journal that (z-y)^(1/n), (z-x)^(1/n) and (x+y)^(1/n) are all...

Wiles' proof is not Fermat's proof, if you believe Fermat's claim to have a proof. Wouldn't it be nice to find a proof that you could teach to a high school algebra class? I tried to use LATEX in my journal. It didn't work. I have been told that the journal will not support LATEX. True or...

"Why would we know that?" If we can agree on the validity of the derivation of the equation (z-y) = k = (x")^n, then, based on symmetry, (z-x) = l = (y")^n and (x+y) = m = (z")^n If we can't agree that (z-y) = k = (x")^n then let's consider the case where n = 3. 3zyk = x^3 - k^3 Since...

I have studied the above for two days and fail to see your point. Perhaps you couild re-phrase it. I fail to see that no matter how you factor (x+y) you cannot prove that (x+y) and (z^n)/(x+y) have common factors, since the latter (z")^n = [(z')^n (z")^n)]/(x+y) and since (x+y)=(z")^n it reduces...

Since I don't claim to be a mathematician, it is quite possible that I have made errors that invalidate the proof that I am attempting. That's why I'm seeking review by mentors such as you. Since k must divide x, x cannot be prime and therefore must have at least two factors greater than 1...

Victor is username "victor.sorokine" who last month posted numerous attempts at proofs of Fermat's Last Theorem. To have access to my journal, just click on "read my journal" under my username. You are right. Science and religion aren't interchangable. Unfortunately, the journal...

FLT-Solution for prime values of "n". The proof is posted in my journal. It has been blessed by two Math academics. Take a look. By the way, Victor was very close.

Thanks ROBPHY. I thought that might be the problem.