1
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ Can you explain how you know this list is Galois-closed? $\endgroup$ – Kevin Casto Oct 30 '20 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. Oct 30 '20 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 Oct 30 '20 at 18:11
  • 5
    $\begingroup$ Evaluate them numerically and expand the product of $(x-x_i)$ in Wolframalpha, for example. $\endgroup$ – Fedor Petrov Oct 30 '20 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 Oct 31 '20 at 20:53
6
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ I got the same result. Thank you very much! +1 $\endgroup$ – Michael Rozenberg Oct 30 '20 at 19:09
  • $\begingroup$ Improved the code. This one is definitely exact (and gives the same answer). $\endgroup$ – darij grinberg Oct 30 '20 at 19:10
2
$\begingroup$

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]

$\endgroup$
1
  • $\begingroup$ I got the same result. Thank you very much! $\endgroup$ – Michael Rozenberg Oct 30 '20 at 19:11
2
$\begingroup$

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)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.