Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
$$T_n(x)=\frac12\left(\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^‌​n\right)$$ might be helpful. –  J. M. Feb 14 '12 at 10:09
    
We generally don't like being ordered to do things here ("Show that..."). We'd be much happier if you put the question in your own words, told us why you found it interesting, showed us what you know about the subject, how far you got, where you got stuck, etc. Oh, and if it's homework, please use the homework tag. –  Gerry Myerson Feb 14 '12 at 11:17
1  
Perhaps the property that $T_{mn}(x) = T_n(T_m(x))$ might be useful. –  marty cohen Feb 15 '12 at 6:00

3 Answers 3

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.

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.