Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $3$.

share|improve this question
11  
About the case when $n$ is odd: $f_n(-1)=0$ hence $x+1$ divides $f_n(x)$, end of the proof. –  Did Nov 15 '11 at 8:16
4  
The document says that it is an open problem. –  Dinesh Nov 15 '11 at 9:28
1  
It is open because it has not yet been determined for all even $n$. –  Cameron Buie May 14 '12 at 21:54
1  
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 $p$. –  A Walker May 9 '13 at 22:18
    
@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. –  Martin Brandenburg May 27 '13 at 21:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.