English

On Schur's irreducibility results and generalised $\phi$-Hermite polynomials

Number Theory 2023-06-06 v1

Abstract

Let cc be a fixed integer such that c{0,2}.c \in \{0,2\}. Let nn be a positive integer such that either n2n\geq 2 or 2n+13u2n+1 \neq 3^u for any integer u2u\geq 2 according as c=0c = 0 or not. Let ϕ(x)\phi(x) belonging to Z[x]\mathbb{Z}[x] be a monic polynomial which is irreducible modulo all primes less than 2n+c2n+c. Let ai(x)a_i(x) with 0in10\leq i\leq n-1 belonging to Z[x]\mathbb{Z}[x] be polynomials having degree less than degϕ(x)\deg\phi(x). Let anZa_n \in \mathbb{Z} and the content of (ana0(x))(a_na_0(x)) is not divisible by any prime less than 2n+c2n+c. For a positive integer jj, if uju_j denotes the product of the odd numbers j\leq j, then we show that the polynomial anu2n+cϕ(x)2n+j=0n1aj(x)ϕ(x)2ju2j+c\frac{a_{n}}{u_{2n+c}}\phi(x)^{2n}+\sum\limits_{j=0}^{n-1}a_j(x)\frac{\phi(x)^{2j}}{u_{2j+c}} is irreducible over the field Q\mathbb{Q} of rational numbers. This generalises a well-known result of Schur which states that the polynomial j=0najx2ju2j+c\sum\limits_{j=0}^{n}a_j\frac{x^{2j}}{u_{2j+c}} with ajZa_j \in \mathbb{Z} and a0=an=1|a_0| = |a_n| = 1 is irreducible over Q\mathbb{Q}. We illustrate our result through examples.

Keywords

Cite

@article{arxiv.2306.01767,
  title  = {On Schur's irreducibility results and generalised $\phi$-Hermite polynomials},
  author = {Anuj Jakhar},
  journal= {arXiv preprint arXiv:2306.01767},
  year   = {2023}
}