Chebyshev Polynomials and Primality Testing 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. 
 A: I think the result for $T$ is in Hong Jen Hsiao, On factorization of Chebyshev's polynomials of the first kind, Bull. Inst. Math. Acad. Sinica 12 (1984), no. 1, 89–94, MR0743938 (86e:11017). Also, it looks like there is a proof in Rayes, M. O.; Trevisan, V.; Wang, P. S.; Factorization properties of Chebyshev polynomials, Comput. Math. Appl. 50 (2005), no. 8-9, 1231–1240, MR2175585 (2007e:33010), a version of which is available at http://icm.mcs.kent.edu/reports/1998/ICM-199802-0001.pdf. 
A: 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.
A: The following was found at http://perso.uclouvain.be/alphonse.magnus/num1a/chebprim.htm
From: Robin Chapman
[1] Re: Primality Test Using Chebyshev Polynomials!
Date: Fri Feb 20 05:35:18 EST 1998

Its roots are the non-zero numbers of the form $\cos(\pi/2n + 2j \pi)$
where $j$ is an integer. They include $\cos(\pi/2n)$ which generates the
real subfield of the cyclotomic field $Q(\exp(2 \pi i/4n))$. This cyclotomic
field has degree $\phi(4n) = 2 \phi(n)$ (where $\phi$ is the Euler function)
and its real subfield has degree $\phi(n)$ [this is due to the
irreducibility of the cyclotomic polynomials]. So the minimal polynomial of
$\cos(\pi/2n)$ has degree $\phi(n)$, and so equals the Chebyshev divided by $x$
iff $\phi(n) = n-1$ iff $n$ is prime.
A related but simpler "primality test" is that p is prime iff
Of course both of these "tests" are useless in practical terms.

Robin Chapman                           "256 256 256.
Department of Mathematics                O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK        12 tyms 256? Bugird if I no.
rjc@maths.exeter.ac.uk                   2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html  Iain M. Banks - Feersum Endjinn
