If $p_n$ is the $n$'th prime, let $A_n(x) = x^n + p_1x^{n-1}+\cdots + p_{n-1}x+p_n$. Is $A_n$ then irreducible in $\mathbb{Z}[x]$ for any natural number $n$?
I checked the first couple of hundred cases using Maple, and unless I made an error in the code those were all irreducible.
I have thought about this for a long time now, and asked many others, with no answer yet.
[Math] Polynomial with the primes as coefficients irreducible
nt.number-theorypolynomials
Best Answer
I will prove that $A_n$ is irreducible for all $n$, but most of the credit goes to Qiaochu.
We have $$(x-1)A_n = b_{n+1} x^{n+1} + b_n x^n + \cdots + b_1 x - p_n$$ for some positive integers $b_{n+1},\ldots,b_1$ summing to $p_n$. If $|x| \le 1$, then $$|b_{n+1} x^{n+1} + b_n x^n + \cdots + b_1 x| \le b_{n+1}+\cdots+b_1 = p_n$$ with equality if and only $x=1$, so the only zero of $(x-1)A_n$ inside or on the unit circle is $x=1$. Moreover, $A_n(1)>0$, so $x=1$ is not a zero of $A_n$, so every zero of $A_n$ has absolute value greater than $1$.
If $A_n$ factors as $B C$, then $B(0) C(0) = A_n(0) = p_n$, so either $B(0)$ or $C(0)$ is $\pm 1$. Suppose that it is $B(0)$ that is $\pm 1$. On the other hand, $\pm B(0)$ is the product of the zeros of $B$, which are complex numbers of absolute value greater than $1$, so it must be an empty product, i.e., $\deg B=0$. Thus the factorization is trivial. Hence $A_n$ is irreducible.