Chebyshev Polynomials Proof Problem

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 polynomials over the rational numbers.

By definition $T_n(x) = \cos(n \cos^{-1}(x))$ . Listing out the first few odd primes of $n$ shows that (for simplicity sake, we take the positive root only)

$T_3(x) - 1 = (4x^3-3x) - 1 = (x-1)(2x+1)^2 \\ \Rightarrow \sqrt{\frac{T_3(x) - 1}{x-1}} = 2x + 1$

$T_5(x) - 1 = (16x^5-20x^3+5x) - 1 = (x-1)(4x^2+2x-1)^2 \\ \Rightarrow \sqrt{\frac{T_5(x) - 1}{x-1}} = 4x^2+2x-1$

$T_7(x) - 1 = (64x^7-112x^5+56x^3-7x) - 1 = (x-1)(8x^3+4x^2-4x-1)^2 \\ \Rightarrow \sqrt{\frac{T_7(x) - 1}{x-1}} = 8x^3+4x^2-4x-1$

$\begin{aligned} T_{11}(x) - 1 &=& (1024 x^{11}-2816 x^9+2816 x^7-1232 x^5+220 x^3-11x) - 1 \\ &=& (x-1) (32 x^5+16 x^4-32 x^3-12 x^2+6 x+1)^2 \\ \Rightarrow \sqrt{\frac{T_{11}(x) - 1}{x-1}} &=& 32 x^5+16 x^4-32 x^3-12 x^2+6 x+1 \end{aligned}$

All these polynomials are irreducible polynomials over the rationals. Is it true for all odd prime $p$ ?

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Comments

The solution is as I expected click here.I hope you don't mind that I copied the problem to math overflow.It uses only the multiplicative property of the degrees of number fields and that $x^{p-1}+x^{p-2}+...+1$ , the pth cyclotomic polynomial, is irreducible (can be proven by shifting x to x+1 and using eisenstein).

Bogdan Simeonov - 6 years ago

Haha, it's a bit of an overkill to use OverFlow! I was about to post this in StackExchange if there's no answer. Thanks!

Pi Han Goh - 5 years, 12 months ago

I posted it on stackexchange, but no one provided a good solution :D

Bogdan Simeonov - 5 years, 12 months ago

If you shift x to x+1 the irreducibility of these can be proven with Eisenstein's criterion.

Thanks! But how do I prove that it's true for ALL odd primes $p$ ?

Pi Han Goh - 6 years ago

By the way do you have a solution or just spotted that they are irreducible for small values?Vecause I can't figure out what identity to use to make that square root in a better form.Eisenstein still works though

I don't have a solution, hence the "or disprove" statement. I got inspired by solving via the Challenge Master notes comment on my solution here.

@Pi Han Goh – So it is perfectly possible that a counterexample exists.Well I'm going to read up on Tschebyscheff Polynomials, any reccomendations?Also, I have found a paper that states that (T_p(x)/x/) is irreducible for all primes p.But I don't know how to deal with that -1.

@Bogdan Simeonov – Recommendation? Haha, no, I read up a lot before posting this, and I'm still pretty stumped. Link please?

@Pi Han Goh – this paper is interesting

I think that the identity $T_n(x)=cos(n.arccosx)$ can give us the roots of $T_n(x)-1$

I'm listening. Please continue on your elaboration.

Well cosx is 1 only when x is 2.pi.something.So that means that $n.arrcosx$ must be 2.pi.something.So $x=cos(2.\pi.k/n)$ are the roots of T_n(x)-1.Problem is, I don't know when they double roots and how to prove that T is their minimal polynomial.It is interesting that these roots are the real parts of the roots of unity.

@Bogdan Simeonov – Also the identity $(T_n(x)-1)/(x-1)(T_n(x)+1)/(x+1)=U_{n-1}$ might help, since we know the factorisation of U.

This post needs to be seen by more people so we can actually solve it lol.Also I still cannot understand why Eisenstein works here (what is special about the coeffs)

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$