TITLE: Chebyshev Polynomials and Primality Testing
QUESTION [5 upvotes]: Let $T(n,x)$ be the nth Chebyshev polynomial of the first kind and let $U(n-1,x)$ the $(n-1)$th Chebyshev polynomial of the second kind. Would any one kindly help show that 
1) $n$ is prime iff $T(n,x)$ is irreducible in $\mathbb{Z}[x]$.
2) $n$ is prime iff $U(n-1,x)$, expressed in powers of $(x^2-1)$, is irreducible in $\mathbb{Z}[x]$. 
Many Thanks!!

No "ordering" is implied here. The wording of the question was done in similar fashion  as any question in any math/research question. I do not see how a person who is asking for help would be ordering people to help. 
Any way back to the topic,
For the first part of the question, I noticed that if n is prime, then T(n,x) satisfies 
Eisenstein's Irreducibility Criterion. But I am not sure how to show if T(n,x) is irreducible then n is prime.

REPLY [2 votes]: I proved the following result, see  http://arxiv.org/abs/1110.6620
Let $\psi_n(x)$ be the minimal polynomial of the algebraic integer $2 \cos \frac{2 \pi}{n}$. Then
$$ U_n(x)=\prod_{\substack{ j|2n+2 \\  j\not=1,2}} \psi_j(2x) \ . $$
Let $n=2^{\alpha} N$ where $N$ is odd and let $r=2^{\alpha+2}$. Then
$$  T_n(x)=\frac{1}{2}\prod_{\substack{ j|N \\   }} \psi_{r j}(2x) \ . $$
For example to prove i) (which should be corrected).  Let $n=p$ prime. Then $\alpha=0$, $N=p$, and $r=4$. 
Therefore
$$ T_n(x)=\frac{1}{2} \psi_4(2x)   \psi_{4p}(2x) $$
Note that $\psi_4(x)=x$. 
As a result,  $n$ prime is equivalent to $\frac{1}{x} T_n(x)$ is irreducible.