Irreducibility of an integer polynomial with exponents in linear sequence? 
Let $b$ and $n$ be two positive integers. Is there are a general result which tell us when the polynomial $$1+x^{b}+x^{2b}+x^{3b}+\cdots+x^{nb}$$
  is irreducible over the integers?

I know that $$1+x+\cdots+x^{n-1}$$
is irreducible if and only if $n$ is prime. I also know that $$1+x^n$$ is irreducible if and only if $n=2^k$ for some $k\geq 0$. However, I'd like to look at a linear sequence and know irreducibility before painstakingly looking for factors.
I've noticed that the $n$ above in the question must be even for otherwise we could factor out $1+x^b$. But other than that I haven't made much ground except for particular values of $n$. I'm kind of hoping for a 'one-fell-swoop' result, but if there isn't, I'd like a recommendation for a general plan of approach for determining irreducibility of these polynomials.
Any help is appreciated.
 A: As pointed out by Bruno Joyal, your polynomial is
$$
f_{b,n}(x)=\frac{1-x^{b(n+1)}}{1-x^b}.
$$
Therefore all the primitive roots of unity of order $b(n+1)$ are zeros of $f_{b,n}(x)$. The minimal polynomial of those roots of unity is the cyclotomic polynomial $\Phi_{b(n+1)}(x)$.
Hence the cyclotomic polynomial is always a factor of $f_{b,n}(x)$. As the cyclotomic polynomials are all irreducible, $f_{b,n}$ is irreducible if and only if it is equal to $\Phi_{b(n+1)}(x)$. On the other hand
$$
\deg\Phi_{b(n+1)}(x)=\phi(b(n+1)).
$$
So a necessary and sufficient condition for the irreducibility of $f_{b,n}(x)$ is that $\phi(b(n+1))=bn$.
But if $n+1$ is not a prime number, then $\phi(n+1)<n$. This implies that there are more than $b$ numbers in the range $1\ldots b(n+1)$ that have a non-trivial common factor with $n+1$, namely those with a remainder modulo $n+1$ that has a common factor with $n+1$. So in that case $\phi(b(n+1))<bn$.
Hence $n+1$ must be a prime number $p$. Now, if $b$ has a prime factor $q$, $q\neq p$, then 
$$\phi(bp)\le bp(1-\frac1q)(1-\frac1p)=bn(1-\frac1q)<bn.$$ On the other hand, if $b=p^k$, then $b(n+1)=p^{k+1}$. In that case $$\phi(b(n+1))=\phi(p^{k+1})=(p-1)p^k=nb.$$ Therefore we have proven 

Answer. The polynomial $f_{b,n}(x)$ is irreducible in $\Bbb{Z}[x]$ if and only if $n+1$ is a prime number $p$, and $b=p^k, k\ge0$.

