TITLE: Are polynomials of the form : $ f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$ irreducible over $\mathbb{Z} $?
QUESTION [16 upvotes]: Is it true that polynomials of the form :

$ f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$
where $\gcd(n+1,k+1)=1$ , $ a\in  \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$

are irreducible over the ring of integers $\mathbb{Z}$?
Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form.
Example :
The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial.
EDIT :
Note that general form for $f_n$ is : $f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $a_1 \neq 1$ is equivalent to the condition $k \geq 1$ . Also polynomial can be rewritten into form :
$$f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$$

REPLY [2 votes]: The case $a=2$ is solved in the following post on my blog: http://mathproblems123.wordpress.com/2009/11/02/irreductible-polynomial/ It is a part of Miklos Schweitzer 2009 contest. A generalization can be found in the second part of the post, generalization which I proposed for the Romanian Team Selection Tests for IMO in 2010.
The polynomial you present clearly satisfies the conditions of the following property: http://mathproblems123.wordpress.com/2009/11/09/position-of-roots/, which shows that the roots of your polynomial have modulus strictly greater than 1 or your polynomial has as a root a root of unity of some order. This might help you derive some properties which prove irreducibility, using roots. 
When $a$ is prime, the method in the first post can be applied and your polynomial is irreducible. When $a$ is composite, things might get trickier.