## Is this a short, marvelous proof of Fermat's Last Theorem?

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

Sending to check the quote fuction.
 Quote by willacaleb 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 he means to press the "QUOTE" button. That way, you will see the raw tex code and can use it as a template and as examples to do your own stuff.
 Recognitions: Homework Help Clicking the "quote" button at the bottom-right of one of my posts should send you to the advanced editor with everything I wrote in quote tags at the top. The math stuff will be inside tex and itex tags. Crash course: So Code: $$\frac{x^2+y^{2-x}}{z_a^b} \neq \sqrt{z^n}$$ renders as $$\frac{x^2+y^{2-x}}{m_a^b} \neq \sqrt{z^n}$$ ... chosen to have everything you'd needed so far in one place. From that you can see: Powers and subscripts work the way you think they do - use curly brackets to group terms. Special terms have a back-slash in front of them and they operate on whatever is inside the curly-brackets after them. i.e. to make a fraction you use \frac{a}{b} for $\frac{a}{b}$ and the square root is sqrt{a} for $\sqrt{a}$ but $\sqrt{ab} \neq \sqrt{a}b$ (\sqrt{ab} \neq \sqrt{a}b) and $x^24 \neq x^{24}$ (x^24 \neq x^{24}) see? There's lots of handy stuff like you get all the Greek characters just by typing their name after a backslash - capitalize the first letter to get the upper-case character. If you want to put an equation inline like that - use the "itex" tags, otherwise use "tex" tags. If you use the $\Sigma$ button in the advanced editor, it will also give you the latex codes. I think there's a quick overview of the PF-Latex codes somewhere - and there are no end of online resources. Now you give it a go ;)

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 Quote by Simon Bridge I think there's a quick overview of the PF-Latex codes somewhere - and there are no end of online resources.
Yes, there is: http://www.physicsforums.com/showthr...=1#post3977517
 Recognitions: Homework Help @micromass: thanks.
 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 think it centers around my 'Case 3' and 'Case 3's' reference to 'Case 1'. I want to respond to each comment in particular and elaborate Case 3. To do that I need more time than my obligations allow at the moment. Again, thank you for your comments and your information. Willacaleb
 A short reply: Last night I think I realized where the sticking point is in many of the comments recieved 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 the assumption of equality in Case 3 requires a congruent number relation, it requires the use of Case 1. Otherwise, we would be working with non-congruent numbers, i.e., 1 < > 1, which would be meaningless. The purpose of, ± u^2, in Cases 2 and 3 is to ensure congruent number units. willacaleb

 Quote by willacaleb A short reply: Last night I think I realized where the sticking point is in many of the comments recieved 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 the assumption of equality in Case 3 requires a congruent number relation, it requires the use of Case 1. Otherwise, we would be working with non-congruent numbers, i.e., 1 < > 1, which would be meaningless. The purpose of, ± u^2, in Cases 2 and 3 is to ensure congruent number units. willacaleb
Yes, but the problem is that case 1 applies ONLY IF $x^2 + y^2 = z^2$ and in case 3 you assume that $x^2 + y^2 > z^2$. So, I don't know where case 1 comes from. Also, I am confuesd about why you expend so much energy in case 3 to deduce that $x^2 + y^2 > z^2$ when this was assumed at the start. Why can't you apply case 1 right then and there?

Also, I am concerned that it took you roughly 20 lines for case 1 and case 2 - which are trivial - and about 5 or so lines for case 3 - which is the hard case. This right there sends up red flags.
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus This has been discussed enough. The proof is not valid. The crucial mistake is refering to Case 1 when proving Case 3, this is not allowed. Thread locked.