In this document on page $3$ I found an interesting polynomial: $$f_n=\frac{(x+1)^n-(x^n+1)}{x}.$$ Question is whether this polynomial is irreducible over $\mathbf{Q}$ for arbitrary $n \geq 1$.

In the document you may find a proof that polynomial is irreducible whenever $n=2p , p \in \mathbf{P}$ and below is my attempt to prove that polynomial isn't irreducible whenever $n$ is odd number greater than $3$. Assuming that my proof is correct my question is :

Is this polynomial irreducible over $\mathbf{Q}$ when $n=2k , k\in \mathbf{Z^+}$ \ $\mathbf{P} $ ?

Lemma 1: For odd $n$, $a^n + b^n = (a+b)(a^{n-1} - a^{n-2} b + \cdots + b^{n-1})$.

Binomial expansion: $(a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \cdots + \binom{n}{n} b^n$. $$\begin{align*} f(x) &= \frac{(x+1)^n - x^n-1}{x} = \frac{(x+1)^n - (x^n+1)}{x} \\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x} \\ &= \frac{(x+1) \cdot (x+1)^{n-1} - (x+1)(x^{n-1}-x^{n-2}+ \cdots + 1)}{x} \\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + 1 \Bigr) - (x^{n-1}-x^{n-2}+ \cdots + 1) \right]}{x} \\ &= \frac{(x+1) \cdot \left[ \Bigl(\binom{n-1}{0}x^{n-1}+ \binom{n-1}{1} x^{n-2} + \cdots + \binom{n-1}{n-2} x \Bigr) - (x^{n-1}-x^{n-2}+ \cdots - x) \right]}{x} \\ &= (x+1) \cdot \small{\left[ \left(\binom{n-1}{0}x^{n-2}+ \binom{n-1}{1} x^{n-3} + \cdots + \binom{n-1}{n-2} \right) - (x^{n-2}-x^{n-3}+ \cdots - 1) \right]} \end{align*}$$

So, when $n$ is an odd number greater than $1$, $f_n$ has factor $x+1$. Therefore, $f_n$ isn't irreducible whenever $n$ is an odd number greater than $3$.

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    $\begingroup$ About the case when $n$ is odd: $f_n(-1)=0$ hence $x+1$ divides $f_n(x)$, end of the proof. $\endgroup$ – Did Nov 15 '11 at 8:16
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    $\begingroup$ The document says that it is an open problem. $\endgroup$ – Dinesh Nov 15 '11 at 9:28
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    $\begingroup$ It is open because it has not yet been determined for all even $n$. $\endgroup$ – Cameron Buie May 14 '12 at 21:54
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    $\begingroup$ When you ask for $f_n$ to be irreducible, does this include constant factors? It appears that the content of $f_n$ is $1$ unless $n=p^k$ is a prime power, in which case the content is $p$. $\endgroup$ – awwalker May 9 '13 at 22:18
  • $\begingroup$ @A. Walker: You are right. It is only interesting to ask for irreducibility over $\mathbb{Q}$. From $f_4 = 2 (2 x^2 + 3 x + 2)$ I infer that actually this was meant, therefore I've edited the question. $\endgroup$ – Martin Brandenburg May 27 '13 at 21:25

For even $n$ we bound the function from below to show it has no real roots and thus no rational roots.

First, note that for $x\ge 0$ $$f_n(x) = \sum_{k=0}^{n-2}{n\choose k+1} x^k = n + \ldots \ge n,$$ where the function expressed by $\ldots$ is clearly positive semidefinite. (We take the sum above to be the definition of $f_n$ as implied by the problem statement, i.e, $f_n(0)=n$.) Thus, the function has no real roots for $x\ge 0$.

One can show that $$f_n(x) = \sum_{j=0}^{(n-2)/2} \frac{n(n-j-2)!}{(j+1)!(n-2j-2)!} (-x)^j (x+1)^{n-2j-2}.$$ For even $n$ and $x\in (-\infty,-1)\cup(-1,0)$ each term in the sum is explicitly positive definite. In addition, $f_n(-1)=2$. Therefore, for even $n$, $f_n$ has no real roots.


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