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Nov 12, 2023 at 5:53 vote accept Er Bu
Nov 7, 2023 at 19:04 comment added paul garrett Btw, really it should be Girard-Newton formulas, to give credit where credit is due. :)
Nov 7, 2023 at 15:56 answer added loup blanc timeline score: 0
Nov 6, 2023 at 22:21 history became hot network question
Nov 6, 2023 at 17:33 answer added Peter Mueller timeline score: 9
Nov 6, 2023 at 16:36 history edited Michael Hardy CC BY-SA 4.0
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Nov 6, 2023 at 15:39 comment added joro I think if $n=3$,$S_4$ doesn't have solutions for all algebraic $a$, only for $a$ root of $a^4 - 6*a^3 + 11*a^2 - 6*a=0$.
Nov 6, 2023 at 15:01 comment added Federico Poloni I have seen the definition, but still I think there is something wrong in the formula for $\sigma_n$. For instance, how many factors are there in the numerator if $a$ is not a positive integer? Say, $a=-1$ or $a = 1+i$?
Nov 6, 2023 at 14:48 comment added Er Bu Sorry, maybe I don't give a clear definition of $\sigma_i$
Nov 6, 2023 at 14:43 history edited Er Bu CC BY-SA 4.0
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Nov 6, 2023 at 14:38 history edited Er Bu CC BY-SA 4.0
added 32 characters in body
Nov 6, 2023 at 14:33 history edited Er Bu CC BY-SA 4.0
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Nov 6, 2023 at 14:33 comment added Federico Poloni Are you sure about the formula for $\sigma_n$? Shouldn't the first factor be $a$, too?
S Nov 6, 2023 at 14:20 review First questions
Nov 6, 2023 at 15:43
S Nov 6, 2023 at 14:20 history asked Er Bu CC BY-SA 4.0