Is this polynomial irreducible over the rationals?

Question

Prove (or disprove): Define $T_n(x)$ as the Chebyshev polynomial of the first kind with degree $n$ . If $p$ is an odd prime, then $\sqrt{\frac{T_p(x)-1}{x-1}}$ is an irreducible polynomial over the rational numbers.

It seems to work shifting $x$ to $x+1$ and using Eisenstein's criterion, but I have no information about the coefficients in the general case. We can use the Pell equation definition and put the polynomial of the second kind to use or maybe use that $T_n(x)=cos(n.arccosx)$ and see when it is equal to $1$, but I have yet to make any progress.

Answer 1

Since $T_p(x)=\cos(p\arccos(x))$, the roots of your polynomial are $\cos(2n\pi/p)=(e^{2n\pi i/p}+e^{-2n\pi i/p})/2$ ($n=1,\dots,(p-1)/2$). It is irreducible, since $\cos(2\pi/p)$ has degree $(p-1)/2$ over $\mathbb{Q}$ ($e^{2\pi i/p}$ has degree $p-1$ over $\mathbb{Q}$ and degree $2$ over $\mathbb{Q}(\cos(2\pi/p))$).