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