Let $Q : \mathbb{Z} \rightarrow \mathbb{Z}$ be a polynomial. Form the set


$$M\_{Q} := \{p:\text{ }p\text{ is prime, }\exists n\_{p}\in \mathbb{Z}\text{ so that }p|Q(n\_{p})\}$$


Is $$\sum\_{s \in M\_{Q}}\frac{1}{s} = \infty ?$$


This question is asked so that one can somewhat understand the density of primes dividing polynomials.


$\textbf{Some special cases:}$


If $Q$ is linear the result is true, if $Q(0) = 0$ the result is true, If $Q(z) = z^2+1$ the result is true.



Divergence of primes dividing polynomials
Yes, the series diverges. We can reduce easily to the case of irreducible monic $Q$.


Next, let $\alpha\_Q(p)$ be the number of roots of $Q(x)$ in $\mathbb{Z}/p\mathbb{Z}$. Note that $M\_Q$ is the set of primes $p$ for which $\alpha\_Q(p)$ is positive. Landau's Prime Ideal Theorem (applied to the field $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of $Q$) tells us that $\alpha\_Q$ is, on average, equal to $1$. That is,
$$\sum\_{p \le x} \alpha\_Q(p) \sim \frac{x}{\log x}.$$
This is because, apart from finitely many primes, the number of ideals in $\mathcal{O}\_{K}$ of norm $p$ coincides with $\alpha\_Q$ by Dedekind's Factorization Theorem. (See [K. Conrad's notes on it](https://kconrad.math.uconn.edu/blurbs/gradnumthy/dedekindf.pdf).)


With partial summation this leads to
$$\sum\_{p \le x} \frac{\alpha\_Q(p)}{p} \sim \log \log x.$$
Since $\alpha\_Q(p) \le \deg Q$, we have that
$$\sum\_{p \le x,\, p \in M\_Q} \frac{1}{p} \ge \frac{1}{\deg Q}\sum\_{p \le x} \frac{\alpha\_Q(p)}{p},$$
yielding the result. (A more detailed discussion, on which this answer was based, is given in p. 36 of "An Introduction to Sieve Methods and Their Applications" by Cojocaru and Murty.)




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Of course, Landau's Theorem is quite an overkill. For you purposes, it suffices to show that the Dedekind zeta function of $K$ (defined above) diverges as $s \to 1^+$, which is a rather simple fact. For instance, see David E. Speyer nice proof [here](https://math.stackexchange.com/a/3274941/10312).


Knowing that, one can mimic Euler's proof of the divergence of $\sum\_p 1/p$ and extend it to the ideal setting from which the answer follows.


One the other hand, $\sum\_{p \in M\_Q} 1/p = \infty$ implies that the Dedekind zeta function of $K$ diverges as $s\to 1^+$. So the question you asked is directly *equivalent* to this divergence.




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Using more sophisticated 'prime number theorems' one can go beyond just $$\sum\_{p \in M\_Q, \, p \le x} 1/p = \Theta( \log \log x)$$
and obtain
$$\sum\_{p \in M\_Q, \, p \le x} 1/p \sim c \log \log x$$
where $c$ is the following rational number: it is the fraction of permutations in the Galois group of the splitting field of $Q$ that have a fixed point. Indeed, by the Frobenius' density theorem we have
$$\sum\_{p \in M\_Q, \, p \le x} 1 \sim c \frac{x}{\log x}$$
for this $c$, and now partial summation can be applied. (See [B Sury's notes on the theorem](https://www.isibang.ac.in/%7Esury/frobreso.pdf). Of course, its analytic aspects can be simplified if we only want the sum of $1/p$.)


