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Let $n \in \mathbb{N}^*$ and $p$ be a prime number. Is the polynomial $f=X^{2n+1}+(p+1)X^{2n}+p \in \mathbb{Z}[X]$ irreducible?

reservoir
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  • What do you think about the problem? – Erick Wong Jul 31 '17 at 19:44
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    I have tried using some classic ideas inspired by this PDF: http://yufeizhao.com/olympiad/intpoly.pdf Unfortunately none of them work in this case (it only "almost" works with Perron). I have tried using similar tricks, but the idea must be different. I currently have no idea for this problem, and my experience regarding irreducibility of polynomials is limited. – reservoir Jul 31 '17 at 19:51
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    Without any context for where this problem came up for you, it's impossible to know, for example, whether Eisenstein's criterion is the kind of thing you're looking for. – Greg Martin Jul 31 '17 at 21:49

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I once remarked in an answer to Show an infinite family of polynomials is irreducible that even in the "equality case" one can use Perron, given that the polynomial $P(z)$ has no zero on the unit circle.

For this you need to use the extended Rouche's theorem (the "Symmetric Version" in https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem ).

Indeed, take $f = P(z)$ and $g = a_{n-1} z^{n-1}$. Then $|f(z)| + |g(z)| > |g(z)| = |a_{n-1}| \ge 1 + |a_{n-2}| + ... + |a_0| \ge |f(z) - g(z)| $ on $|z| = 1$.

So you know that $P(z)$ has $n-1$ roots (strictly) inside the unit circle. Proceed as in the article you quoted from now on to show that $P(z)$ is irreducible.

Apply this here.

Shubhodip Mondal
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