tag:blogger.com,1999:blog-12535639.post113052381487025932..comments2018-03-17T17:58:54.013-07:00Comments on Fermat's Last Theorem: Fermat's Last Theorem: Proof for n=5Larry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-12535639.post-71756919415918057272015-07-04T15:17:09.536-07:002015-07-04T15:17:09.536-07:00(4^12 + 7^5)^6 =(16,794,023)^6 translates into (4^...(4^12 + 7^5)^6 =(16,794,023)^6 translates into (4^12 * 16,794,023)^5 + (7^5 * 16,794,023)^5 = (16,794,023)^6 which can also be expressed to the power 5. So I am not sure why Fermat's Last Theorem, which is easy to prove false, ceases to be of interest, since it helps with the ABC conjecture proof. I do not have a calculator to even check my math, plus I am very calculator inept when it some to anything beyond basic math, but even if there is some other factorial, this works. Debra AxonDebra Kay Axonhttps://www.blogger.com/profile/16328941512036956388noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-10389948446573327752013-11-12T09:46:11.088-08:002013-11-12T09:46:11.088-08:00Hi Larry, concerning:
x5 + y5 = z5 → xyz = 0 if x...Hi Larry, concerning:<br /><br />x5 + y5 = z5 → xyz = 0 if x,y,z are integers<br /><br />could you expand on what the arrow means.<br /><br />Thanks<br />KeithKeith Griffithshttps://www.blogger.com/profile/10801471962320796343noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-54411060902097262262009-10-16T18:07:28.121-07:002009-10-16T18:07:28.121-07:00Proof of Case I for n = 5
The following is quite e...Proof of Case I for n = 5<br />The following is quite elegant. I have not seen this elsewhere.<br /><br />The trinomial expansion of (x+y+z)^5 simplifies down to<br /><br />(x+y+z)^5 = 5(z+x)(z+y)(x+y){x^2+y^2+z^2+xy+xz+yz}<br /><br />assuming that x^5 + y^5 + z^5 = 0<br /><br />this expression can be neatly rearranged as<br /><br />= 5 (z+x)(z+y)(x+y)*<br /> {(x+y)^2+(z+y)^2+(z+x)^2}/2<br /><br />fascinating isn’t it ?<br /><br />In mod 5, all non 0 squares are either 1 or 4 <br /><br />(1^2 = 1,2^2 = 4,3^2 = 4,4^2 = 1)<br /><br />and the sum of 3 squares (of non zero numbers) in mod 5 can never be 0 <br /><br />(they must sum to 1, 2, 3 or 4).<br /><br />The possible combinations are; <br />(1+1+1) = 3, <br />(1+1+4) = 1, <br />(1+4+4) = 4 and <br />(4+4+4) = 2<br /><br />Therefore including (z+x)(z+y)(x+y) which also cannot be divisible by 5 under Case I, <br /><br />the whole expression on the right hand side can contain only one prime factor of 5.<br /><br />This makes the left hand side expression impossible as all primes in the expression must be raised to the 5th power, at least and 5 is only present once on the right hand side.<br /><br />Therefore <br />(x+y+z)^5 = 5 (z+x)(z+y)(x+y)* {(x+y)^2+(z+y)^2+(z+x)^2}/2 <br /><br />is impossible for case I, where 5 does not divide x, y or z and x^5 + y^5 + z^5 = 0<br /><br />This completes the proof of Case I for n = 5neat_mathshttps://www.blogger.com/profile/09923473294000429170noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-3645234723154109802008-10-10T13:45:00.000-07:002008-10-10T13:45:00.000-07:00Hi Pau,I am very glad to hear that my blog is help...Hi Pau,<BR/><BR/>I am very glad to hear that my blog is helping. <BR/><BR/>Cheers,<BR/><BR/>-LarryLarry Freemanhttps://www.blogger.com/profile/06906614246430481533noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-10973134468669094422008-10-10T13:29:00.000-07:002008-10-10T13:29:00.000-07:00Hi Larry. My name is Pau and I'm from Spain. I'm d...Hi Larry. My name is Pau and I'm from Spain. I'm doing 2nd of baccalaureate and also i'm doing a research work called "Fermat's last theorem". I'm doing it with a classmate and your blog is helping us a lot.<BR/>Thank you very much, Larry.<BR/>A cordial greeting from Spain.Pauhttps://www.blogger.com/profile/10935192193836139570noreply@blogger.com