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Define complex conjugate as the pair of complex numbers $a+bi,a-bi$. Assume you have 5 such pairs and they are the 10 roots $x_i$ of a 10th-deg equation with integer coefficients. In general, how many combinations,

$$F= x_1 x_2+x_3 x_4+x_5 x_6+x_7 x_8+x_9 x_{10}$$

will be real, and how do we tell Mathematica to enumerate them all?

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  • $\begingroup$ Is this a puzzle? $\endgroup$
    – Dr. belisarius
    Nov 13, 2013 at 5:06
  • $\begingroup$ No, I'm wondering if the behavior of the PSL(2,5) sextic and PSL(2,7) octic generalizes to higher degrees. That is, I'll use Recognize to determine if there is an F such that it is an algebraic number of degree one less than the $x_i$, just like what happens for the cited equations with those Galois groups. $\endgroup$
    – Tito Piezas III
    Nov 13, 2013 at 5:22
  • $\begingroup$ NB: Recognize is now RootApproximant (since version 6 or 7) $\endgroup$
    – Dr. belisarius
    Nov 13, 2013 at 5:26
  • $\begingroup$ I'm using a very old version. V4. <sheepish> $\endgroup$
    – Tito Piezas III
    Nov 13, 2013 at 5:27
  • $\begingroup$ For the generic case I think you need to have either 1, 3, or 5 of the products involve a pair of conjugates, with pairs of the rest using swapped conjugates e.g. zw*+zw, where I use ' for complex conjugate, not multiplication (and multiplication is just juxtaposition). If I am correct, and if I also wrote down the combinatorics correctly, then the number of combinations should be 5*(6*4)*(3*2) + 5*4*(3*2) + 5*4*3 = 900. $\endgroup$
    – Daniel Lichtblau
    Nov 13, 2013 at 15:22

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