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How to find roots of $X^5 - 1$? (Or any polynomial of that form where $X$ has an odd power.)

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    $\begingroup$ You mean complex roots? Just write an arbitrary complex number as $re^{i\theta}$, substitute into the polynomial and solve first for $r$, then for $\theta$. $\endgroup$
    – Alex B.
    May 15, 2011 at 12:29
  • $\begingroup$ Note that the comment of Alex B applies to any polynomial of the form $x^n-c$ where $n$ is a positive integer and $c$ is a complex number. $\endgroup$ May 15, 2011 at 23:35
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    $\begingroup$ Over which domain? In other words, what is the type of $X$? Does $X \in \mathbb{Z}$ or $\in \mathbb{C}$ etc? $\endgroup$
    – user2468
    May 16, 2011 at 16:28
  • $\begingroup$ @M.S., I think we've all been assuming what's wanted is solutions in the complex numbers, and Paul hasn't raised any objections, so it's safe to conclude that that's the intention. I can't imagine anyone needing help to find the roots in the integers. $\endgroup$ May 16, 2011 at 23:19

3 Answers 3

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It is easy to write down the solution in trigonometric functions, namely, $x=e^{2\pi i n/5}$ for $n=0,\ldots, 4$. Here is an algebraic solution. First factor the polynomial as $$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$ The first factor gives you the obvious solution $x=1$. To find other roots, we need to solve the equation $$ x^4+x^3+x^2+x+1=0.$$ Divides both sides by $x^2$, and use the fact that $(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$, we get $$ (x+\frac{1}{x})^2 +(x+\frac{1}{x})-1=0$$ Set $y=x+\frac{1}{x}$, and we get a quadratic equation $$ y^2+y-1=0,$$ which one may easily solve using quadratic formulas. Once we have solved for $y=y_0, y_1$, the equation reduces into two other quadratic equations $$ x+ \frac{1}{x}= y_i, \qquad i=0, 1 $$ or equivalently, $$ x^2-y_i x+1=0.$$ We just apply the quadratic formula again.

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    $\begingroup$ @Jiangwei: I like your solution :) $\endgroup$
    – user9413
    May 15, 2011 at 12:53
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    $\begingroup$ Very nice. But Paul asked about "any polynomial of that form where $x$ has an odd power." If you took $x^{11}-1$, you'd run into something a lot worse than quadratics. The trig functions are, of course, still available. $\endgroup$ May 15, 2011 at 23:32
  • $\begingroup$ Very elegant and nice. +1 $\endgroup$ May 16, 2011 at 12:12
  • $\begingroup$ @Gerry: Trig functions equivalent to a complex variable as in Chandru's answer or of other kind? $\endgroup$ May 16, 2011 at 12:17
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    $\begingroup$ @Gerry: I thought you had in mind something different. So $e^{i\frac{2\pi (1+k)}{n}}$, $k=0,1,\ldots ,n-1$ are the solutions of $x^{n}-1=0$. $\endgroup$ May 16, 2011 at 13:59
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Edit 2: Since the equation $x^n-1=0$ is equivalent to $x=\sqrt[n]{1}$, it has the following $n$ solutions:

$$e^{i\dfrac{2\pi (1+k)}{n}}=\cos\left( \dfrac{2\pi (1+k)}{n}\right)+i\sin\left( \dfrac{2\pi (1+k)}{n}\right),\qquad k=0,1,\ldots ,n-1,$$

where $n$ is a positive integer, odd or even.

See comment by Gerry Myerson indicating how the Galois group for an irreducible quartic determines whether the method exposed below works.


Although late I present another algebraic solution, less elegant but more "automatic", in the sense that no clever trick is necessary. [Edit in response to Gerry Myerson's comment] We are going to factor it into two factors, one being linear and the other quadratic. This method works for quintic polynomials, as long as we are able to find one solution and solve the quadratic polynomial factor. If it fails, in principle we can solve the quadratic polynomial by means of a resolvent cubic equation (also in this post in Portuguese) [end of edit]. For the general polynomial of 7th degree or higher it is not applicable, because it depends on finding one root by inspection (or numerically) and the remaining ones algebraically.

By inspection we see that $x=1$ is a zero of $x^{5}-1$. By long division or Ruffini's rule, we find

$$x^{5}-1=\left( x-1\right) \left( x^{4}+x^{3}+x^{2}+x+1\right) .$$

Now we factor $x^{4}+x^{3}+x^{2}+x+1$ into two quadratic polynomials

$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) ,$$

whose (real) coefficients have to be found by comparing LHS with the expanded RHS. Since

$$\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right)$$

$$ =x^{4}+\left( B+b\right) x^{3}+\left( C+bB+c\right) x^{2}+\left(bC+cB\right) x+cC,$$

the coefficients must satisfy

$$\left\{ \begin{array}{c} B+b=1 \\ C+bB+c=1 \\ bC+cB=1 \\ cC=1% \end{array}% \right. \Leftrightarrow \left\{ \begin{array}{c} B=1-b \\ 1/c+b\left( 1-b\right) +c=1 \\ b/c+c\left( 1-b\right) =1 \\ C=1/c% \end{array}% \right. $$

One of the solutions of $b/c+c\left( 1-b\right) =1$ is $c=1$. By substitution we find the remaining coefficients:

$$\left\{ \begin{array}{c} B=\frac{1}{2}\mp \frac{1}{2}\sqrt{5} \\ b=\frac{1}{2}\pm \frac{1}{2}\sqrt{5} \\ c=1 \\ C=1% \end{array}% \right. $$

Thus

$$x^{4}+x^{3}+x^{2}+x+1=\left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5}% \right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}% \right) x+1\right) $$

and finally

$$x^{5}-1=\left( x-1\right) \left( x^{2}+\left( \frac{1}{2}+\frac{1}{2}\sqrt{5% }\right) x+1\right) \left( x^{2}+\left( \frac{1}{2}-\frac{1}{2}\sqrt{5}% \right) x+1\right) .$$

Hence the roots of $x^{5}-1$ are the roots of these three factors. They are, respectively, $x_1=1$ and

$$x_{2,3}=-\frac{1}{4}-\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10+2\sqrt{5}},\quad x_{4,5}=-\frac{1}{4}+\frac{1}{4}\sqrt{5}\pm \frac{1}{4}\sqrt{-10-2\sqrt{5}}.$$

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  • $\begingroup$ How general is this "general method"? It's not going to work for $x^{11}-1$, is it, and that would seem to be the kind of polynomial Paul has in mind. $\endgroup$ May 15, 2011 at 23:30
  • $\begingroup$ @Gerry Myerson: See the revised answer. $\endgroup$ May 15, 2011 at 23:36
  • $\begingroup$ @Americo, thanks. I suspect that as a "general method for quintic polynomials, as long as we are able to find one solution" it's going to run into some difficulties with, say, $(x-1)(x^4-x-1)$. $\endgroup$ May 16, 2011 at 0:03
  • $\begingroup$ @Gerry: That's right. Applying this method to $(x^{4}-x-1)=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) $ I was left with a system of equations, one of which is $-1/c-c^{2}/\left( 1+c^{2}\right) ^{2}+c=0$. $\endgroup$ May 16, 2011 at 10:15
  • $\begingroup$ ... Does Abstract Algebra provides some criteria to know in advance if a quartic polynomial can be factored in this way? I know nothing of Abstract Algebra. Anyway, one can solve quartic equations by a resolvent cubic equation. I didn't check if we run into troubles when applied to $x^{4}-x-1=0$. $\endgroup$ May 16, 2011 at 10:15
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Hint: Use De Moivre's theorem and note that $$1^{1/5} = \Bigl( \cos 2m\pi + i \sin{2m\pi} \Bigr)^{1/5} \ ;\qquad m \in \mathbb{Z}$$

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