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.



Polynomial with the primes as coefficients irreducible?
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.


