tag:blogger.com,1999:blog-12535639.post113082353826540271..comments2024-02-26T06:55:41.876-08:00Comments on Fermat's Last Theorem: Fermat's Last Theorem: n=5: 5 divides zLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-12535639.post-263385146417546772010-08-26T03:38:02.666-07:002010-08-26T03:38:02.666-07:00In step (8) should:
u' = c + 5d^2
be
u'...In step (8) should:<br /><br /><i>u' = c + 5d^2</i><br /><br />be<br /><br /><i>u' = c<b>^2</b> + 5d^2</i><br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-75328007731800502562010-08-26T03:32:11.055-07:002010-08-26T03:32:11.055-07:00In step (6e) should
(e) t=a - 5b^2 is a fifth pow...In step (6e) should<br /><br /><i>(e) t=a - 5b^2 is a fifth power since:</i><br /><br />be<br /><br /><i>(e) t=a<b>^2</b> - 5b^2 is a fifth power since:</i><br /><br />as in step (5d)?<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-80831393005822029312010-08-26T03:25:56.937-07:002010-08-26T03:25:56.937-07:00Apologies for the last post.
Step (10) is similar...Apologies for the last post.<br /><br />Step (10) is similar to step (7) and uses the same lemma. <br /><br />Right?<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-61501405425426000372010-08-25T03:49:05.726-07:002010-08-25T03:49:05.726-07:00In step (10):
(10) With the properties in step #9...In step (10):<br /><br /><i>(10) With the properties in step #9, we can use a lemma <b>which I will prove later</b> to show:</i><br /><br />Is the proof of this lemma up yet?<br /><br />:-(<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1425329462235950272010-08-25T03:42:49.916-07:002010-08-25T03:42:49.916-07:00In step (9d) should:
(d) u' - 5v'2 is a f...In step (9d) should:<br /><br /><i>(d) u' - 5v'2 is a fifth power since:</i><br /><br />be<br /><br /><i>(d) u'<b>^2</b> - 5v'2 is a fifth power since:</i><br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-70040030183220847792010-08-25T03:40:50.668-07:002010-08-25T03:40:50.668-07:00In step (9d) should:
(2*5^2r)^2 = 2*5^3*10r^2 = (...In step (9d) should:<br /><br /><i>(2*5^2r)^2 = 2*5^3*10r^2 = (2*5^3)*v</i><br /><br />be<br /><br /><i>(2*5^2r)^2 = 2*5^3*10r^2 = (2*5^3)*<b>b</b></i><br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-2694527623942056842010-08-25T03:31:49.001-07:002010-08-25T03:31:49.001-07:00In step (9d) should
Now, (2*5^2r) is a fifth powe...In step (9d) should<br /><br /><i>Now, (2*5^2r) is a fifth power (#7e)</i><br /><br />be<br /><br /><i>Now, (2*52r) is a fifth power <b>(#6e)</b></i><br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-78545862214219277452010-06-08T07:45:24.428-07:002010-06-08T07:45:24.428-07:00I think in step (6a):
Since f divides a, we know ...I think in step (6a):<br /><br />Since f divides <b>a</b>, we know that f divides 10r^2<br /><br />Should be:<br /><br />Since f divides <b>b</b>, we know that f divides 10r^2<br /><br />RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-13138023142116820802008-05-01T00:28:00.000-07:002008-05-01T00:28:00.000-07:00Hi Tim,Thanks for your comment. I've updated step...Hi Tim,<BR/><BR/>Thanks for your comment. I've updated step #6e. You are right that the detail was not clear.<BR/><BR/>Please let me know if you have questions about any other steps. <BR/><BR/>Cheers,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-21699838534684145092008-04-30T22:37:00.000-07:002008-04-30T22:37:00.000-07:00I think you are missing a square on the a in 6e or...I think you are missing a square on the a in 6e or else i don't make the conection. Love the proof though makes it real nice and simple to follow thank you<BR/><BR/>TimTimhttps://www.blogger.com/profile/17142742150370007410noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-29286622915237436212008-04-30T22:35:00.000-07:002008-04-30T22:35:00.000-07:00I think you left out the square on a in 6e, or els...I think you left out the square on a in 6e, or else i don't follow the proof. Other than that i love the lay out real easy to follow<BR/><BR/>Thanks<BR/><BR/>TimTimhttps://www.blogger.com/profile/17142742150370007410noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-73742953803268793012007-09-30T06:18:00.000-07:002007-09-30T06:18:00.000-07:00Hi Rob,Thanks for the comments! I have tried to r...Hi Rob,<BR/><BR/>Thanks for the comments! I have tried to rewrite the proof to make it clearer.<BR/><BR/>There was an errant minus. It has been removed.<BR/><BR/>The argument in (6d) is confusing. I have changed it to make it clearer.<BR/><BR/>Cheers,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-49025583158129701612007-09-28T07:11:00.000-07:002007-09-28T07:11:00.000-07:00Step (3):Is there an errant minus at the start of:...Step (3):<BR/>Is there an errant minus at the start of:<BR/>-2^5m*5^5n*z'^5 = 2p(p^4 + 10p^2q^2 + 5q^4) ?<BR/><BR/>Step (6d):<BR/>f changes into p and back again.<BR/><BR/>RobScouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.com