I have been trying to solve this problem on my own for four days now, and I cannot figure out how to prove it:
If we express a prime $p$ in base $10$ as $$p= a_m10^m+a_{m-1}10^{m-1}+\ldots +a_110+a_0,$$ with $0\leq a_i \leq 9$ for all $i \in [m]$, then the polynomial $$f(x)= a_mx^m+a_{m-1}x^{m-1}+\ldots +a_1x+a_0$$ is irreducible in $\mathbb{Z}[x]$.
First, note that $\gcd(a_n, \ldots, a_0)=1$, otherwise $p$ would not be prime.
I want to prove this by contradiction. Suppose, to the contrary, that $f(x)$ is reducible in $\mathbb{Z}[x]$. Then there exist $a(x), b(x) \in \mathbb{Z}[x]$, neither of which are units, such that $f(x)=a(x)b(x)$. Now $p=f(10)=a(10)b(10)$, which implies that $a(10)\in \{1, -1\}$ or $b(10)\in \{1, -1\}$ (whichever is the case will imply that the other is $\pm p$). Without loss of generality, suppose that $a(10)=1$.
Now, I was given the following hint: Show that $f(x)$ takes on infinitely many prime values, that is, show that show that there exists a sequence $\{x_n\}$ of integers such that $f(x_n)=q_n$, where $q_n$ is prime for each $n$. If I can show this, I am done because by the argument made above, without loss of generality, this would imply that $a(x)$ takes on the value 1 infinitely many times. Then $a(x)-1=0$ for infinitely many $x$-values, but since a polynomial has infinitely many zeros if and only if it is the zero function itself, this would imply that $a(x)=1$ for all integers $x$, a contradiction.
Can someone help me figure out why the hint is true? Thank you!! This problem is very interesting, don't you agree?