# How do you factor x^11+y^11?

May 10, 2015

${x}^{11} + {y}^{11} = \left(x + y\right) \left({x}^{10} - {x}^{9} y + {x}^{8} {y}^{2} - {x}^{7} {y}^{3} + {x}^{6} {y}^{4} - {x}^{5} {y}^{5} + {x}^{4} {y}^{6} - {x}^{3} {y}^{7} + {x}^{2} {y}^{8} - x {y}^{9} + {y}^{10}\right)$

First notice that if $y = - x$ then ${y}^{11} = {\left(- x\right)}^{11} = - {x}^{11}$. So $\left(x + y\right)$ is a factor. The other factor is easy to find, requiring alternating $\pm 1$ coefficients to cause the intermediate terms to cancel out.

If we were allowed complex number coefficients, there would be a complete factorization as follows:

${x}^{11} + {y}^{11} = \left(x + y\right) \left(x + \tau y\right) \left(x + {\tau}^{2} y\right) \left(x + {\tau}^{3} y\right) \left(x + {\tau}^{4} y\right) \left(x + {\tau}^{5} y\right) \left(x + {\tau}^{6} y\right) \left(x + {\tau}^{7} y\right) \left(x + {\tau}^{8} y\right) \left(x + {\tau}^{9} y\right) \left(x + {\tau}^{10} y\right)$

where $\tau$ stands for the complex 11th root of 1.

Clearly, none of the factors apart from $\left(x + y\right)$ has real coefficients, so there are definitely no other linear factors with rational coefficients.

I think that if any strict subset of the complex linear factors is multiplied together, we get a polynomial with at least one complex coefficient.

UPDATE:
Actually ${\tau}^{10} = \overline{\tau}$, ${\tau}^{9} = \overline{{\tau}^{2}}$, etc.
So $\left(x + \tau\right) \left(x + {\tau}^{10}\right) = \left({x}^{2} + \left(\tau + \overline{\tau}\right) x + 1\right)$ does have real coefficients.

$\tau = \cos \left(\frac{2 \pi}{11}\right) + i \sin \left(\frac{2 \pi}{11}\right)$

and $\tau + \overline{\tau} = 2 \cos \left(\frac{2 \pi}{11}\right)$

etc.

May 26, 2015

Here's a fairly terse, advanced version...

Let $\tau = \cos \left(\frac{2 \pi}{11}\right) + i \sin \left(\frac{2 \pi}{11}\right)$

$\tau$ is a primitive complex 11th root of unity.

Then for any $n \in \mathbb{Z}$ we have

${\tau}^{n} = \cos \left(\frac{2 n \pi}{11}\right) + i \sin \left(\frac{2 n \pi}{11}\right)$

and:

${\tan}^{11 - n} = \cos \left(2 \pi - \frac{2 n \pi}{11}\right) + i \sin \left(2 \pi - \frac{2 n \pi}{11}\right)$

$= \cos \left(- \frac{2 n \pi}{11}\right) + i \sin \left(- \frac{2 n \pi}{11}\right)$

$= \cos \left(\frac{2 n \pi}{11}\right) - i \sin \left(\frac{2 n \pi}{11}\right)$

$= \overline{{\tau}^{n}}$

So

${\tau}^{n} + \overline{{\tau}^{n}} = 2 \cos \left(\frac{2 n \pi}{11}\right)$

Now

${x}^{11} + {y}^{11} = {\prod}_{n = 0}^{n = 10} \left(x + {\tau}^{n} y\right)$

$= \left(x + y\right) {\prod}_{n = 1}^{n = 10} \left(x + {\tau}^{n} y\right)$

$= \left(x + y\right) {\prod}_{n = 1}^{n = 5} \left(\left(x + {\tau}^{n} y\right) \left(x + \overline{{\tau}^{n}} y\right)\right)$

$= \left(x + y\right) {\prod}_{n = 1}^{n = 5} \left({x}^{2} + \left({\tau}^{n} + \overline{{\tau}^{n}}\right) x y + {y}^{2}\right)$

$= \left(x + y\right) {\prod}_{n = 1}^{n = 5} \left({x}^{2} + 2 \cos \left(\frac{2 n \pi}{11}\right) x y + {y}^{2}\right)$