tag:blogger.com,1999:blog-6883415296937284014.post1769831747467778058..comments2016-06-09T23:11:19.160-07:00Comments on Religion, Sets, and Politics: Fermat's Little Theorem, Euler's Generalization and Artin RepresentationsJoshuahttp://www.blogger.com/profile/00637936588223855248noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6883415296937284014.post-57505259303794491212013-08-22T02:59:05.707-07:002013-08-22T02:59:05.707-07:00This is gorgeous!This is gorgeous!Hollishttp://besttreadmillforhomes.us/noreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-66765557444225153882013-07-22T20:32:18.346-07:002013-07-22T20:32:18.346-07:00Juan, there's no conjectured asymptotic. Other...Juan, there's no conjectured asymptotic. Other work by Ambrose shows that the growth is at least x^(1.7) (actually an ugly constant a little above 1.7). Joshuahttp://www.blogger.com/profile/00637936588223855248noreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-23202186426301696482013-07-19T09:12:42.912-07:002013-07-19T09:12:42.912-07:00I made some experiments (taking a = 7)
it appears ...I made some experiments (taking a = 7)<br />it appears that G(x) x^(-2)log^4 x still decreases. <br /><br />With G(x) x^(-1.5) log^b x it appears to depend on b. decreasing <br />with b=0 but increasing for b=2. <br /><br />These results are with modest values x = 100000.<br /><br />Are there some conjectures about the <br />asymptotic behavior of G(x)? <br /><br />(The statement of Euler Theorem has <br />a typo, p should be n).<br /><br />JuanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-73975501360139363902013-07-12T17:07:40.693-07:002013-07-12T17:07:40.693-07:00Yes, thanks.Yes, thanks.Puzzledhttp://www.blogger.com/profile/10022350351268826353noreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-79199629888126304552013-07-12T07:55:14.243-07:002013-07-12T07:55:14.243-07:00Puzzled, the thing is that Fermat's theorem on...Puzzled, the thing is that Fermat's theorem only applies to primes. If is prime, then n not dividing a is equivalent to (n,a)=1. But this isn't the same condition in the composite case, since for example, 4 doesn't divide 6 but (4,6)=2. So we need the stronger condition there. Does that make sense? Joshuahttp://www.blogger.com/profile/00637936588223855248noreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-76391499483603514932013-07-10T05:45:55.923-07:002013-07-10T05:45:55.923-07:00Sorry, rereading my comment I don't like the t...Sorry, rereading my comment I don't like the tone. I meant to say that I was confused by it. In any case, I think I'm still a bit confused. In Fermat's theorem, we required that n not divide a. In Euler's, the requirement is that (n,a)=1 - do we call this a strengthening since, if n divided a, the gcd would be as large as possible, so gcd=1 is an extreme form of not dividing, so to speak? Not only does it not divide, but also they share nothing at all (dividing would mean, I guess, that they share all of n's prime factors?)Puzzledhttp://www.blogger.com/profile/10022350351268826353noreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-88951131230976734382013-07-09T20:55:40.786-07:002013-07-09T20:55:40.786-07:00Puzzled, it is called I made a typo. That should b...Puzzled, it is called I made a typo. That should be n not dividing a in the first bit. Does that make sense? Joshuahttp://www.blogger.com/profile/00637936588223855248noreply@blogger.comtag:blogger.com,1999:blog-6883415296937284014.post-6276214707597439942013-07-09T20:47:20.755-07:002013-07-09T20:47:20.755-07:00>First, one needs to strengthen the condition o...>First, one needs to strengthen the condition on a. So instead of n|a, >one wants that a and n have greatest common divisor 1. <br /><br />How is this a strengthening?Puzzledhttp://www.blogger.com/profile/12866127197554237039noreply@blogger.com