Recent content by willacaleb

A short reply: Last night I think I realized where the sticking point is in many of the comments received to date: Is it valid in Case 3 to use Case 1? I say it is valid because Case 1 is the only congruent number relation, i.e., 1 = 1, of the three relations of the Trichotomy Law. And since...

Thanks very much for the information regarding LaTex. Because of several obligations I am going to spend a few days formulating what I hope is a full response to each comment and question to date. I have been thinking about the general area of concern expressed in most of the responses and I...

Sending to check the quote fuction.

Simon Bridge: Your conversions are helpful and accurate. I would like to use LaTex, however, when I hit the "quote button" nothing happens. I am in "Quick Reply." I will send the answers I wrote today tomorrow. The site kicked them out when I tried to send them. willacaleb

I think this will clarify my attempt at proving FLT: Let or assume, x^2 + y^2 < = > z^2, and assume, x^n + y^n = z^n, n>2. Then, by the fundamental operation of dividing equals-by-equals, x^n/z^n-2 + y^n/z^n-2 = z^n/z^n-2 = z^2, would be true for the assumption we have made for the < = >...

Simon Bridge, Abacus, Robert1986, Micromass: I am pressed for time today but I will respond tomorrow. Thank you and I appreciate your responses. Willacaleb

Simon Bridge: Yes, Case 1 demonstrates the singular case that pythagorean triples cannot satisfy FLT. Case 2 and Case 3 demonstrate that the other possible triples when, x^2 + y^2 < > z^2, cannot satisfy FLT. These three cases exhaust all possible whole numbers. Sorry about the 'primitive'...

Given: x, y, z, n are whole natural numbers, including the regular and irregular prime numbers; x and y < z. The, "Trichotomy Law", which states there are three relations possible between (x^2 + y^2 ) and z^2, i.e., < = > . Any square can be subdivided into an infinite number of 'sum of two...