prove that $x^{20}+(1-x)^{20}-20$ is square free

Question

Prove that $x^{20}+(1-x)^{20}-20$ is square free (i.e. has no repeated roots).

Note that the claim holds iff $p'(x)$ is coprime to $p(x)$, where $p(x)=x^{20}+(1-x)^{20}-20$. $p'(x)= 20x^{19}-20(1-x)^{19}.$ Also, $\gcd(p,p') = \gcd((1-x)^{19}-20, x^{19}-20)$ and the roots of $x^{19}-20$ are $\sqrt[19]{20} \exp (2\pi i k /19)$ for some $0\leq k\leq 18$ while the roots of $(1-x)^{19}-20$ are $1-\sqrt[19]{20} \exp (2\pi i k/19)$ for some $0\leq k\leq 18$. I'm not sure how to proceed from here.

Answer 1

The problem has been reduced to show that $\gcd((1-x)^{19}-20, x^{19}-20)=1$ in $\Bbb C[x]$.

Since both $(1-x)^{19}-20$ and $x^{19}-20$ are in $\Bbb Q[x]$, it is enough to prove $\gcd((1-x)^{19}-20, x^{19}-20)=1$ in $\Bbb Q[x]$.

Let us work in $\Bbb Q[x]$ from now on.

Since $5\mid 20$, $5\not\mid1$ and $5^2\not\mid20$, Eisenstein's criterion tells us $x^{19}-20$ is irreducible. In other words, $x^{19}-20$ is a prime polynomial. Any polynomial that shares a non-constant factor with $x^{19}-20$ must either be a constant multiple of it or have a higher degree. Since $(1-x)^{19}-20$ is neither, it is coprime to $x^{19}-20$.

Answer 2

First, the roots of $x^{19}-20$ are $20^{\frac{1}{19}}\exp(\frac{2\pi ik}{19})$ for $k=0,\cdots,18$, and the roots of $(1-x)^{19}-20$ are $1-20^{\frac{1}{19}}\exp(\frac{2\pi i \ell}{19})$ for $\ell=0,\cdots,18$. In order to show that the desired gcd is 1, we only need to show that there is no common root of the two polynomials. In other words, there are no $k,\ell\in\mathbb{Z}$ such that $$ 20^{\frac{1}{19}}\exp(\frac{2\pi ik}{19}) = 1-20^{\frac{1}{19}}\exp(\frac{2\pi i\ell}{19}), $$ or equivalently, $$ \exp(\frac{2\pi ik}{19})+\exp(\frac{2\pi i\ell}{19})=20^{-\frac{1}{19}}. $$ Note that $20^{-\frac{1}{19}}$ is a real number, and for any $z_1,z_2\in\mathbb{C}$, if $z_1+z_2\in\mathbb{R}$, then the imaginary parts of $z_1$ and $z_2$ must cancel. Thus, we only need to consider the case where $\ell=-k$ (since the 19th roots of unity are not symmetric about the y-axis). Then the above reduces to $$ \cos(\frac{2\pi k}{19})=\frac{1}{2}\cdot 20^{-\frac{1}{19}}. $$ Clearly $k\not\in 19\mathbb{Z}$. Then the degree of $\cos(\frac{2\pi k}{19})$ is 9, but the degree of $\frac{1}{2}\cdot 20^{-\frac{1}{19}}$ is 19. This shows that there cannot exist $k\in\mathbb{Z}$ such that the above holds, as desired.