tag:blogger.com,1999:blog-12535639.post112546406896619432..comments2018-03-17T17:58:54.013-07:00Comments on Fermat's Last Theorem: Sophie's ProofLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-12535639.post-91423054936048070562017-12-09T07:48:41.778-08:002017-12-09T07:48:41.778-08:00Hi
I have a question concerning step 5 and 10
in ...Hi<br /><br />I have a question concerning step 5 and 10<br />in step 5.d you have <br />y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1) ≡<br />y^(n-1) - y^(n-2)(-y) + ... -y(-y)^(n-2) + (-y)^(n-1) (mod p)<br />which you rewrite as <br />y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1) ≡(n)y^(n-1) (mod p)<br /><br />and then you write that p either divides n or y^(n-1)<br />but i thought since it is a modular equation, that the abovestanding meant that p divides<br />(y^(n-1) - y^(n-2)z + ... - yz^(n-2) + z^(n-1))-((n)y^(n-1) <br /><br />that is the left side minus the right side of the equation<br /><br /><br />I also dont quite get where the information about x^n+y^n+z^n≡ 0 (mod 2n+1) comes from in step 10.c<br /><br />Isabella Sofie Kofoedhttps://www.blogger.com/profile/15052352095111387131noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-6801035524529500302016-12-16T20:20:29.003-08:002016-12-16T20:20:29.003-08:00Hi Annika,
This blog is not very active any more ...Hi Annika,<br /><br />This blog is not very active any more but I still monitor comments and occasionally answer comments. :-)<br /><br />The idea of setting z' to -z is about introducing another variable.<br /><br />Let's say we have x^n + y^n = z^n where n=1 (which has a solution. We need n to be odd for this to work).<br /><br />In this case: 3^1 + 4^1 = 7^1 <br /><br />Let us define a new variable which I will call z'= -z (but I could also call it w -- it's math so the name doesn't matter so much).<br /><br />Since z' = -z, it follows that z' = -7.<br /><br />Now we can say that:<br /><br />3^1 + 4^1 + (-7)^1 = 0<br /><br />So, it follows that x^n + y^n + (z')^n = 0<br /><br />Once it is clear that we can do this, we can for all purposes, analyze the problem of x^n + y^n = z^n in the form of x^n + y^n + z^n = 0. Yes, each time we say z, we really mean z' but since this is clear, just to keep it cleaner, I only refer to it as z.<br /><br />What I probably should have done to make this point clear is to have a separate lemma where I write:<br /><br />If x^n + y^n = z^n has a solution then there exists a,b,c where a^n + b^n + c^n = 0 (and a = x, b = y, c = -z)<br />Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-34464102701954875302016-12-15T12:24:42.667-08:002016-12-15T12:24:42.667-08:00Is this blog still active?
I'm doing a project...Is this blog still active?<br />I'm doing a project for school, and I'm having a hard time understanding how you set z' to -z in order to have the equation: x^n+y^n+(z')^n=0<br />But when you move on with the proof you leave out the prime-symbol. Why is this?Annika Lundhttps://www.blogger.com/profile/09594418285137253919noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-76716042644424679442009-12-11T14:51:36.024-08:002009-12-11T14:51:36.024-08:00Hey. ... I dont understand somethin. Sophie Germai...Hey. ... I dont understand somethin. Sophie Germains proof of FSS, was that n was a prime larger than 2 but less than 100.<br />but this proof doesnt say anything about that ?or?thullehttps://www.blogger.com/profile/02590617786012451579noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-26673608235992101852007-09-30T17:20:00.000-07:002007-09-30T17:20:00.000-07:00Hi Rob,You are right on both counts. I have made ...Hi Rob,<BR/><BR/>You are right on both counts. I have made the changes to the blog. <BR/><BR/>Thanks for noticing! :-)<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-6075635530201736762007-09-27T05:22:00.000-07:002007-09-27T05:22:00.000-07:00Step (10c):There is an errant zero:x^n + y^n + (-z...Step (10c):<BR/>There is an errant zero:<BR/>x^n + y^n + (-z)^n0 ≡ 0 (mod 2n+1)<BR/>Instead of:<BR/>x^n + y^n + (-z)^n ≡ 0 (mod 2n+1)<BR/><BR/>Step (14a)(i):<BR/>2n+1 divides b^(2n+1)<BR/>Should be:<BR/>2n+1 divides b^n ?Scouse Robhttps://www.blogger.com/profile/00144454830208958210noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143958042300839022006-04-01T22:07:00.000-08:002006-04-01T22:07:00.000-08:00Just fixed it. Thanks.-LarryJust fixed it. Thanks.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143928905206171162006-04-01T14:01:00.000-08:002006-04-01T14:01:00.000-08:00step 3 I think it should sayx^5+y^5-(-z')^5 or x^5...step 3 I think it should say<BR/>x^5+y^5-(-z')^5 or x^5+y^5+(z')^5broghttps://www.blogger.com/profile/08721338360491222656noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143927118110151292006-04-01T13:31:00.000-08:002006-04-01T13:31:00.000-08:00On which step, do you see a mistake in the sign? ...On which step, do you see a mistake in the sign? I will be glad to change it if there is a typo.<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143925766681993842006-04-01T13:09:00.000-08:002006-04-01T13:09:00.000-08:00yeah - I've worked through it now - I think there ...yeah - I've worked through it now - I think there is just a sign wrong on one line of the proof but the rest seems right.broghttps://www.blogger.com/profile/08721338360491222656noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143923464521336412006-04-01T12:31:00.000-08:002006-04-01T12:31:00.000-08:00The idea here is that:z^5 = -(-z)^5This is done so...The idea here is that:<BR/>z^5 = -(-z)^5<BR/><BR/>This is done so that we can change:<BR/>x^5 + y^5 = z^5<BR/><BR/>into<BR/><BR/>x^5 + y^z + [-(-z)^5] = 0<BR/><BR/>If this is what you meant, then you are correct.<BR/><BR/>The reason for this is so that we have a symmetric equation of the form:<BR/><BR/>x^5 + y^5 + z^5 = 0<BR/><BR/>In this symmetric equation, we can make an argument about x^5 and know that it could just as well apply to y^5 and z^5.Larry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-1143922376630401682006-04-01T12:12:00.000-08:002006-04-01T12:12:00.000-08:00it says here that you set z' to -z so x^5+y^5=(-z'...it says here that you set z' to -z so x^5+y^5=(-z')^5 and so x^5+y^5-(-z')^5 or am I mistaken?broghttps://www.blogger.com/profile/08721338360491222656noreply@blogger.com