Consider the Chebyshev polynomial of the first kind $T\_n(x)$ and its factorization in $\mathbb F\_p$ for a given prime $p$. Most often, this factorization is not *uniform* (meaning that the irreducible factors have not all the same degree, as in [this](https://mathoverflow.net/questions/191049/which-criteria-for-uniformly-splitting-polynomials) thread). But the following seems to hold: 



> 
> For given $n$, $T\_n(x)$ factors uniformly for all $p$ iff $n$ is a power of $2$. 
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More precisely: define $\nu(p):=\max( \nu\_2(p-1),\nu\_2(p+1))=\nu\_2(p^2-1)-1$ (where as usual, $\nu\_2(\cdot)$ is the multiplicity of the prime factor $2$).  

Then it appears that each irreducible factor of $T\_{2^k}(x)$ over $\mathbb F\_p$ has degree $2^{max(k+2-\nu(p),0)}$.


How to prove this?



Chebyshev polynomials factoring uniformly modulo all primes
Let $\zeta$ be a primitive $4n$th root of unity. Then $\alpha = (\zeta+\zeta^{-1})/2$ is a root of $T\_n$. $\alpha$ generates the real subfield of
the $4n$th cyclotomic field, of degree $\varphi(4n)/2$ over $\mathbb Q$
(where $\varphi$ is the Euler phi function). Now $\varphi(4n) \le 2n$, with
equality if and only if $n$ is a power of 2. This implies that $T\_n$ is irreducible
over $\mathbb Q$ if and only if $n$ is a power of 2. The uniformity follows
from that (since the field generated by a root is Galois, even cyclic).


Let $n = 2^k$. The degree of an irreducible factor mod $p$ is the smallest $e$ 
such that $p^e \equiv \pm 1 \bmod 2^{k+2}$. If $l = v(p)$, then this is 1 for $k \le l-2$ and $2^(k+2-l)$ for $k \ge l-2$ as claimed.


