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Can the polynomial $$y^2(1-x)^4+x^2(1-y)^4=xy(1-x)^2(1-y)^2(n^2+2)$$ be solved over the roots of unity for an integer $n\ge 3$? Known solutions are $$n=1:\quad x=\zeta_5\quad y=\zeta_5^2$$ $$n=2:\quad x=\zeta_8\quad y=\zeta_8^3$$

As the problem is posed, $x$ and $y$ do not have to necessarily be roots of unity of the same degree. But as far as I know, that may be a necessary condition for solvability.

This is a reformulation of an earlier question Rational Trig Solutions for $n\ge 3$ I was advised to formulate the question as it is presented here because there are apparently strong results on what polynomials can be solved over the roots of unity. There are a few papers on such solutions that pop up on a naif google search, but all have been too dense for me to understand.

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Performing a brute force search for solutions of the form

$$x=\zeta_j \quad y=\zeta_j^k$$

reveals that real solutions for $n$ abound, but integer real solutions for $n$ are not likely for solutions of this form.

In addition to your solutions, I found these solutions $$n = 1: \quad x=\zeta_5 \quad y=\zeta_5^3$$ $$n = 2: \quad x=\zeta_8 \quad y=\zeta_8^5$$ $$n = 9.0009: \quad x=\zeta_{33} \quad y=\zeta_{33}^{11} \quad or \quad y=\zeta_{33}^{22}$$ $$n = 10.007: \quad x=\zeta_{34} \quad y=\zeta_{34}^{13} \quad or \quad y=\zeta_{34}^{21}$$

Here is the simple MatLab/Octave script I used to do the brute force search:

for k = [2:40]
  k
  x = exp(1i*2*pi/k);
  for j = [2:k]
    y = x^j;
    n = sqrt(((y*(1-x)^2-x*(1-y)^2)^2)/(y*(1-x)^2*x*(1-y)^2))
  endfor
endfor
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