nt.number theory - 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.

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