Question

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

Answer 1

`x^11+y^11=(x+y)(x^10-x^9y+x^8y^2-x^7y^3+x^6y^4-x^5y^5+x^4y^6-x^3y^7+x^2y^8-xy^9+y^10)`

First notice that if `y = -x` then `y^11 = (-x)^11 = -x^11`. So `(x+y)` is a factor. The other factor is easy to find, requiring alternating `+-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=(x+y)(x+tau y)(x+tau ^2y)(x+tau ^3y)(x+tau ^4y)(x+tau ^5y)(x+tau ^6y)(x+tau ^7y)(x+tau ^8y)(x+tau ^9y)(x+tau ^10y)`

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

Clearly, none of the factors apart from `(x+y)` 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 = bar tau`, `tau^9 = bar (tau^2)`, etc.
So `(x+tau)(x+tau^10) = (x^2+(tau + bar tau)x+1)` does have real coefficients.

`tau = cos ((2pi)/11) + i sin ((2pi)/11)`

and `tau + bar tau = 2 cos ((2pi)/11)`

etc.

Answer 2

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

Let `tau = cos((2pi)/11) + i sin((2pi)/11)`

`tau` is a primitive complex 11th root of unity.

Then for any `n in ZZ` we have

`tau^n = cos((2npi)/11) + i sin((2npi)/11)`

and:

`tan^(11-n) = cos(2pi-(2npi)/11) + i sin(2pi-(2npi)/11)`

`= cos(-(2npi)/11) + i sin(-(2npi)/11)`

`= cos((2npi)/11) - i sin((2npi)/11)`

`= bar(tau^n)`

So

`tau^n + bar(tau^n) = 2cos((2npi)/11)`

Now

`x^11 + y^11 = prod_(n=0)^(n=10) (x + tau^n y)`

`=(x+y) prod_(n=1)^(n=10) (x + tau^n y)`

`=(x+y) prod_(n=1)^(n=5) ((x + tau^n y) (x + bar (tau^n) y))`

`=(x+y) prod_(n=1)^(n=5) (x^2 + (tau^n + bar(tau^n)) xy + y^2)`

`=(x+y) prod_(n=1)^(n=5) (x^2 + 2 cos ((2npi)/11) xy + y^2)`