tag:blogger.com,1999:blog-12535639.post113662154500121788..comments2017-03-09T11:25:26.636-08:00Comments on Fermat's Last Theorem: Pell's Equation: The Solution with Reduced Quadratic EquationsLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-12535639.post-71667915827398107522010-06-11T02:36:06.722-07:002010-06-11T02:36:06.722-07:00In step (11g) of Theorem 2
This gives us Q-q is ...In step (11g) of <b>Theorem 2</b><br /><br /><br />This gives us Q-q is greater than<br />-2(2+3)=-2(5)=-0.4. <br /><br />should be<br /><br />This gives us Q-q is greater than<br />-2<b>/</b>(2+3)=-2<b>/</b>(5)=-0.4.Scouse Robhttp://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-25952491084274903052010-06-11T02:26:37.748-07:002010-06-11T02:26:37.748-07:00In step (11a) of Theorem 2
From #6, we know that....In step (11a) of <b>Theorem 2</b><br /><br />From <i>#6</i>, we know that..<br /><br />should be<br /><br />From <b>#3</b>, we know thatScouse Robhttp://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-39407536368491774862010-06-11T02:20:53.944-07:002010-06-11T02:20:53.944-07:00In step (4) of Theorem 2
(x-by)/2*(x+by)/2-(ay)(-...In step (4) of <b>Theorem 2</b><br /><br />(x-by)/2*(x+by)/2-(ay)(-cy)= x^2-b^2y^2+acy^2<br /><br />should be<br /><br />(x-by)/2*(x+by)/2-(ay)(-cy)= <b>(x^2-b^2y^2)/4</b>+acy^2<br /><br />RobScouse Robhttp://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-31614235427201215372010-06-11T02:13:54.085-07:002010-06-11T02:13:54.085-07:00Larry
I'm sorry I still don't understand ...Larry<br /><br />I'm sorry I still don't understand <b>Theorem 2</b>.<br /><br /><br /><i>We know that there exists a reduced quadratic equation that has a discriminant=D</i><br /><br />D has to be non-square to be a discriminant?<br /><br />D also has to be equal to 1 or 0 mod 4?<br /><br />Does this not restrict the statement of the Theorem to certain values of d.<br /><br />I am confused.<br /><br />Any help at all with this issue would be appreciated.<br /><br />RobScouse Robhttp://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-14982992485356039422010-06-11T00:40:12.287-07:002010-06-11T00:40:12.287-07:00Good morning Larry.
A new day and a new outlook.
...Good morning Larry.<br /><br />A new day and a new outlook.<br /><br />It seems so clearer now.<br /><br />D=b^2-4ac≡b^2 mod 4.<br /><br />If b is even then D≡0 mod 4 and D=4n.<br /><br />If b is odd then D≡1 mod 4 and D=4n+1.<br /><br />I'm sorry if I have been sending you too many comments over the last few days.<br /><br />I truly think your blog is the greatest thing I have found on the internet (after online poker and, perhaps, metamath.) <br /><br />Thank you for taking the time and the effort of posting all these wonderful proofs, and for working out the details to arrange them into easy to follow step by step proofs.<br /><br />RobScouse Robhttp://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-64021213203420274072010-06-10T09:25:24.609-07:002010-06-10T09:25:24.609-07:00In the link of step (2) of Theorem 2
It is proven...In the link of step (2) of <b>Theorem 2</b><br /><br />It is proven that if D is of the form 4n or 4n+1 then a reduced quadratic equation that has a discriminant = D exists.<br /><br />What about the cases when D=4n+2 or D=4n+3?<br /><br />I'm tired and cannot work it out.<br /><br />I'll have to wait here until tomorrow.<br /><br />Any help in understanding this would be greatly appreciated.<br /><br />Thanks<br /><br />RobScouse Robhttp://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1137988428799152222006-01-22T19:53:00.000-08:002006-01-22T19:53:00.000-08:00I'm amazed that anyone would attempt this. Some on...I'm amazed that anyone would attempt this. Some one that can carry through explaining the whole theory up to and including the final proof of FLT is an incredibly accomplished amateur. I'm inspired to knuckle down and follow the details of the proofs for the 5th and 7th degrees.<BR/><BR/> I'm just a far more rudimentry amateur, who enjoys inventing his own. I hit this site while searching on Pell's equation. I was thinking there should exist a more direct solution, rather than finding the numbers of a continued fraction first (of obscure meaning, if any), and then using those numbers to calculate convergents. I haven't seen anything direct like that on the web, although I haven't looked thoroughly through what I have downloaded yet.Kenneth Florekhttp://www.blogger.com/profile/08362730891375773255noreply@blogger.com