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Fix a nonzero integer $a$ and a positive integer $k$. I'm looking for some criterion to establish for which nonzero integers $b$ and $n \geq 2$ the polynomial $$f(x) := x^n - (x+a)^{n-k} (x+b)^k$$ is irreducible over $\mathbb{Z}[x]$. I think it is irreducible at least for all sufficiently large $n$, but I failed to prove it.

EDIT: The question is more complicated than I expected. So I specify that I'd be happy just to prove that there are infinitely many $n$ such that $f(x)$ is irreducible in $\mathbb{Z}[x]$ for any nonzero integer $b \neq -a$. Peter Mueller's comments below show that these $n$ must be relatively prime with $k$, so maybe a good guess is to consider $n$ prime.

Thank you in advance for any idea.

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    $\begingroup$ In the special case where $a=b$, the roots of your polynomial are the $\zeta a/(1-\zeta)$ where $\zeta\neq 1$ is a $n$-th root of $1$. So the polynomial is irreducible if and only if $n$ is prime. In particular, $n$ large is not sufficient. $\endgroup$
    – Aurel
    Nov 4, 2013 at 11:39
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    $\begingroup$ Whenever $k$ and $n$ are not relatively prime, your polynomial is reducible. $\endgroup$ Nov 4, 2013 at 11:39
  • $\begingroup$ @Peter: not quite "whenever": $n=2$, $k=1$, $a=1$, $b=-1$. $\endgroup$ Nov 4, 2013 at 11:41
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    $\begingroup$ One way to approach this for arbitrary $n,b$ is to pick a prime $p$ dividing $a$ (assuming $a\ne\pm1$), and compute the inertia group of a prime over $p$ in the splitting field of $f(x)$. This might work out nicely because $f(x)$ reduces mod $p$ to $x^{n-k}(x^k-(x+b)^k)$, whose factorization is under control, and then hopefully Hensel's lemma yields the factorization over $\mathbf{Z}_p$. The hope is that these inertia groups will generate a transitive subgroup of $S_n$, implying irreducibility. $\endgroup$ Nov 4, 2013 at 12:22
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    $\begingroup$ Here is a cheap answer to your edited question. Under Artin's conjecture on primitive roots, you can prove there are infinitely many suitable $n$ for each $b\neq a\pm 1$, using the mod $p$ method from my second comment: pick $p\mid b-a$, under Artin's conjecture there are infinitely many primes $n$ such that $p$ is a primitive root modulo $n$, and for those $n$, $f$ is irreducible mod $p$. $\endgroup$
    – Aurel
    Nov 4, 2013 at 22:39

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