tag:blogger.com,1999:blog-12535639.post113082901816681374..comments2024-02-26T06:55:41.876-08:00Comments on Fermat's Last Theorem: Fermat's Last Theorem: n=5 : 5 doesn't divide zLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-12535639.post-32776520496301006012010-08-26T06:31:09.805-07:002010-08-26T06:31:09.805-07:00It took a bit of work to figure out step (12).
Is...It took a bit of work to figure out step (12).<br /><br />Is it because of the following?<br /><br />5^2r is a power of 5 from step (7d, 7e)<br /><br />b=5r^2<br /><br />5^3b = 5^4r^2 = (5^2r)^2<br /><br />So 5^3b is a 5th power.<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-68677274840326874022010-08-26T06:09:17.084-07:002010-08-26T06:09:17.084-07:00In step (9) should:
5^3a
be
5^3b
RobIn step (9) should:<br /><br /><i>5^3<b>a</b></i><br /><br />be<br /><br /><i>5^3<b>b</b></i><br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.com