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I am looking for a seventh degree polynomial with integer coefficients, which has the following roots. $$x_1=2\left(\cos\frac{2\pi}{43}+\cos\frac{12\pi}{43}+\cos\frac{14\pi}{43}\right),$$ $$x_2=2\left(\cos\frac{6\pi}{43}+\cos\frac{36\pi}{43}+\cos\frac{42\pi}{43}\right),$$ $$x_3=2\left(\cos\frac{18\pi}{43}+\cos\frac{22\pi}{43}+\cos\frac{40\pi}{43}\right)$$ $$x_4=2\left(\cos\frac{20\pi}{43}+\cos\frac{32\pi}{43}+\cos\frac{34\pi}{43}\right),$$ $$x_5=2\left(\cos\frac{10\pi}{43}+\cos\frac{16\pi}{43}+\cos\frac{26\pi}{43}\right),$$ $$x_6=2\left(\cos\frac{8\pi}{43}+\cos\frac{30\pi}{43}+\cos\frac{38\pi}{43}\right)$$ and $$x_7=2\left(\cos\frac{4\pi}{43}+\cos\frac{24\pi}{43}+\cos\frac{28\pi}{43}\right).$$ I see only that $\sum\limits_{k=1}^7x_k=-1$, but the computations for $\sum\limits_{1\leq i<j\leq7}x_ix_j$ and the similar are very complicated by hand and I have no any software besides WA, which does not help.

Thank you for your help!

Update.

I got $$\sum\limits_{1\leq i<j\leq7}x_ix_j=-18.$$

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    $\begingroup$ Can you explain how you know this list is Galois-closed? $\endgroup$
    – Kevin Casto
    Commented Oct 30, 2020 at 17:59
  • $\begingroup$ 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 $x_5, x_6, x_7$.) $\endgroup$
    – Jack L.
    Commented Oct 30, 2020 at 18:04
  • $\begingroup$ @Jack L. I got these roots by using a primitive root modulo 43, which is $3$. I fixed a typo. Thank you! $\endgroup$
    – Michael Rozenberg
    Commented Oct 30, 2020 at 18:11
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    $\begingroup$ Evaluate them numerically and expand the product of $(x-x_i)$ in Wolframalpha, for example. $\endgroup$
    – Fedor Petrov
    Commented Oct 30, 2020 at 18:34
  • $\begingroup$ @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. $\endgroup$
    – Michael Rozenberg
    Commented Oct 31, 2020 at 20:53

3 Answers 3

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In SageMath, you can enter the following:

U.<zeta> = CyclotomicField(43)
P.<x> = PolynomialRing(U)

def c(j):  # cos(j * pi / 43)
    return (zeta ** j + zeta ** (-j))/2

x1 = 2*(c(2) + c(12) + c(14))
x2 = 2*(c(6) + c(36) + c(42))
x3 = 2*(c(18) + c(22) + c(40))
x4 = 2*(c(20) + c(32) + c(34))
x5 = 2*(c(10) + c(16) + c(26))
x6 = 2*(c(8) + c(30) + c(38))
x7 = 2*(c(4) + c(24) + c(28))

(x-x1)*(x-x2)*(x-x3)*(x-x4)*(x-x5)*(x-x6)*(x-x7)

And you get:

x^7 + x^6 - 18*x^5 - 35*x^4 + 38*x^3 + 104*x^2 + 7*x - 49

that is: $x^{7} + x^{6} - 18 x^{5} - 35 x^{4} + 38 x^{3} + 104 x^{2} + 7 x - 49$.

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  • $\begingroup$ I got the same result. Thank you very much! +1 $\endgroup$
    – Michael Rozenberg
    Commented Oct 30, 2020 at 19:09
  • $\begingroup$ Improved the code. This one is definitely exact (and gives the same answer). $\endgroup$
    – darij grinberg
    Commented Oct 30, 2020 at 19:10
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By PARI / GP I get

$x^7 + x^6 - 18*x^5 - 35*x^4 + 38*x^3 + 104*x^2 + 7*x - 49$ :

K = nfinit (subst(polcyclo(43),x,y))

w = Mod(y,K.pol)

f0(k) = (w^k + 1/w^k)

f(k1,k2,k3) = f0(k1) + f0(k2) + f0(k3)

v = [f(1,6,7),f(3,18,21),f(9,11,20),f(10,16,17),f(5,8,13),f(4,15,19),f(2,12,14)]

/*

=

[x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49,

x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49,

x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49,

x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49,

x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49,

x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49,

x^7 + x^6 - 18x^5 - 35x^4 + 38x^3 + 104x^2 + 7*x - 49]

*/

mps = [minpoly(w) | w<-v]

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  • $\begingroup$ I got the same result. Thank you very much! $\endgroup$
    – Michael Rozenberg
    Commented Oct 30, 2020 at 19:11
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I also used PARI/GP with the following program:

z1 = Mod(z, (z^43-1)/(z-1));
e(n) = lift(Mod(3,43)^n);
c(n) = z1^n + z1^-n;
r(n) = c(1*n) + c(6*n) + c(7*n);
p = prod(n=1,7, x - r(e(n)));
lift(p)

with the resulting output

z^7+z^6-18*z^5-35*z^4+38*z^3+104*z^2+7*z-49

A simpler program with complex numbers is

z1=exp(2*Pi*I/43); z2=z1^6; z3=z1^7;
bestappr(prod(n=1,7, m=lift(Mod(3,43)^n);\
x - 2*real(z1^m + z2^m + z3^m)), 10^9)
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