Skip to main content
12 events
when toggle format what by license comment
Oct 31, 2020 at 20:53 comment added Michael Rozenberg @Geoff Robinson I think much more interesting to solve the equation $x^7+x^6-18x^5-35x^4+38x^3+104x^2+7x-49=0$, without any hint. For this thing exactly I created it.
Oct 31, 2020 at 20:20 answer added Somos timeline score: 2
Oct 31, 2020 at 8:20 vote accept Michael Rozenberg
Oct 30, 2020 at 19:08 answer added jjcale timeline score: 2
Oct 30, 2020 at 19:07 answer added darij grinberg timeline score: 6
Oct 30, 2020 at 18:38 history edited Michael Rozenberg CC BY-SA 4.0
added 64 characters in body
Oct 30, 2020 at 18:34 comment added Fedor Petrov Evaluate them numerically and expand the product of $(x-x_i)$ in Wolframalpha, for example.
Oct 30, 2020 at 18:11 comment added Michael Rozenberg @Jack L. I got these roots by using a primitive root modulo 43, which is $3$. I fixed a typo. Thank you!
Oct 30, 2020 at 18:08 history edited Michael Rozenberg CC BY-SA 4.0
edited body
Oct 30, 2020 at 18:04 comment added Jack L. Is there a rule for obtaining the (rational) entries of the cosine terms; for the first three, I recognize that the next entries are thrice the previous (modulo $43$). Could you also kindly check if the remaining ones are correct (and I suppose the last three roots are supposed to be $x_5, x_6, x_7$.)
Oct 30, 2020 at 17:59 comment added Kevin Casto Can you explain how you know this list is Galois-closed?
Oct 30, 2020 at 17:31 history asked Michael Rozenberg CC BY-SA 4.0