tag:blogger.com,1999:blog-5376628436133716219.post4016814601261355203..comments2024-03-28T14:45:46.850-07:00Comments on Why I hate physics: Excel To The RescueMarty Greenhttp://www.blogger.com/profile/17624084719249673373noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-5376628436133716219.post-7520169232801103882014-01-04T14:23:32.234-08:002014-01-04T14:23:32.234-08:00Oops...you must have had two typos then. You meant...Oops...you must have had two typos then. You meant to start with x1^2 -6, right? And then you meant that n=8 is the first one that IS separable, right?Marty Greenhttps://www.blogger.com/profile/17624084719249673373noreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-18005488155389039802014-01-04T13:48:20.104-08:002014-01-04T13:48:20.104-08:00No, that is a typo. I AM working over A^2 - n = B,...No, that is a typo. I AM working over A^2 - n = B, B^2 - n = C and C^2 - n = A. I just mistyped it in the post above.Balarkanoreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-43998785353490191792014-01-04T13:34:19.972-08:002014-01-04T13:34:19.972-08:00Careful...sometimes there can be less than meets t...Careful...sometimes there can be less than meets the eye. I haven't thought about this carefully, but you're working a recursive relation that links A^2 not to B but to B^2. Just remember...if A, B and C are roots of a cubic, then A^2, B^2, and C^2 are also roots of a cubic, albeit a different one. (You can see it must be true because the symmetric polynomials in A^2 etc. are readily expressible in terms of the elementary symmetric polynomials.)<br /><br />So are you finding equations in which the three roots differ by integers?? Or am I completely missing something?Marty Greenhttps://www.blogger.com/profile/17624084719249673373noreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-1607007535577602222014-01-04T11:40:48.949-08:002014-01-04T11:40:48.949-08:00This is the case I am looking for now! The octic g...This is the case I am looking for now! The octic generated by xi^2 - 2 = x(i+1)^2 is<br /><br />x^8 - 24x^6 + 204x^4 - 720x^2 - x + 894<br /><br />and after excluding the trivial roots by synthetic division, this reduces to the sextic<br /><br />x^6 + x^5 - 17x^4 - 11x^3 + 91x^2 + 25x - 149<br /><br />... which is not separable over integers anymore. And it has an unusual Galois group: the direct product of C9 and C2. (although admittedly I didn't calculate it by myself)<br /><br />After further effort on GP, I found out that almost all such sextic that is not separable over Z[x] has Galois group C9xC2. In fact, the next n after 4 such that the sextic formed of the cyclic relationships of 3 roots xi^2 - n = x(i+1)^2 is not seperable over Z[x] is 8. Apparently it looks like the powers of 2 would work, but I found out I was fooled by the strong law of small numbers. It turned out the sequence is 2, 4, 8, 14, 22. Definitely something fishy is going around here. I think the latter wouldn't be hard to analyze but what about the Galois group problem?<br /><br />Seems more than the eyes meet.Balarkanoreply@blogger.com