tag:blogger.com,1999:blog-12535639.post113662154500121788..comments2024-02-26T06:55:41.876-08:00Comments on Fermat's Last Theorem: Pell's Equation: The Solution with Reduced Quadratic EquationsLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger6125tag: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 Robhttps://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 Robhttps://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 Robhttps://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 Robhttps://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 Robhttps://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 Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.com