Eisenstein's criterion over polynomials irreducible

Question

If $2n+1$ is a prime number implies that the polynomial $\displaystyle P(x)=\sum_{k=0}^{n}(-1)^{k}{2n+1\choose 2k+1}x^{2k}$ is irreducible over $\Bbb Q$?

Answer 1

Yes, $P(x)$ is irreducible over $\Bbb Q$ since

• the leading coefficient of $P(x)$ is unitary ($(-1)^n$)
• the prime number $2n+1$ divides ${2n+1\choose 2k+1}$ for $k = 0,\dots,n-1$
• when $k = 0$, the constant term ${2n+1\choose 1} = 2n+1$ is not divisible by $(2n+1)^2$

By Einsenstein's Criterion, $P(x)$ is irreducible over $\Bbb Q$.