Sets of Prime Numbers Generated By an Irreducible Monic Polynomial Given a non-constant integral irreducible monic polynomial $f(x)$, the prime factors of its value at integers $x\in\mathbb{N}$ forms a set $\mathcal{P}(f)$.
Is it possible that $\mathcal{P}(f)\cap\mathcal{P}(g)=\emptyset$ for two such polynomials $f$ and $g$.
 A: The answer is no. See Theorem 3 in Prime Numbers in Certain Arithmetic Progressions:

If $f,g \in \mathbb{Z}[x]$ are non-constant, then $P(f) \cap P(g)$ is infinite.

The proof there is quite short so I might as well include it:

Suppose $\alpha$ is a root of $f$ and $\beta$ is a root of $g$. By Dedekind 's theorem, except for a finite number of exceptions, $p$ is a prime divisor of $f$ exactly when $p$ has a first degree prime ideal factor in the field $\mathbb{Q}(\alpha)$. A similar statement holds for $g$. Consider the field, $\mathbb{Q}(\alpha, \beta)=\mathbb{Q}(\theta)$ for some $\theta \in \mathcal{O}_K$. If $h$ is the minimal polynomial of $\theta$ then Schur's Theorem assures us that there are infinitely many primes which have a first degree prime ideal factor in $\mathbb{Q}(\alpha, \beta)$. These primes thus have a first degree prime ideal factor in both $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$. Thus, except for a finite number of exceptions, these primes are in both $P(f)$ and $P(g)$.

A: Update: See Sii’s answer. The conclusion here is incorrect, and there is a fallacy in my reasoning here somewhere.
Yes, it is. Some number theoretic machinery is required to see this as this is far from obvious.
To give a counter example I will make use of Multiplicative Binary Quadratic forms and a  fact about quadratic forms.
A binary quadratic form is an equation of the form $Q(x,y)=ax^2+by^2$, with $a,b$ both nonzero. Binary quadratic forms with $a=1$ and $|b|\in \mathbb{N}$ are always multiplicative. See Ken Ono's paper http://tiny.cc/vk27ix, this is explained in the first paragraph.
That is if $Q(x_1,x_2)=pq$ for some $p,q\in \mathbb{Z}$, then $Q(x_1,x_2)=Q(x_3,x_4)Q(x_5,x_6)$. Of course, all the $x_i$ are integers.
The idea is more general but since you only want one variable functions we will fix both $b,y$. 
Theorems:
$p=x^2+3y^2$ if and only if $ p≡1(mod 3) $
$p=x^2+y^2$ if and only if $ p≡1(mod 4) $
First one is a Fermat conjecture(Euler proved it) and the second he proved, see http://tiny.cc/gu27ix or check out David A. Cox's book "Primes of the form $x^2+ny^2$".
Let 
$Q_1(x,y)=x^2+3y^2$ 
$Q_2(x,y)=x^2+y^2$
Because of the multiplicative property all we need to do is find two quadratic forms where the possible primes set of all prime $p$ generated by each form are either: 
1) $<Q_1(x,y)>_p \cap <Q_2(x,y)>_p={\emptyset}$ $\; \; \; \; \; \;\forall x,y \in \mathbb{Z}$ 
or
2) $<Q_1(x,y_0)>_p \cap <Q_2(x,y_0)>_p={\emptyset}$ $\; \; \; \; \; \;\forall x \in \mathbb{Z}$ , where $y_0$ is a fixed integer.(We can fix $x$ and vary $y$, whichever is easier to get.)
Clearly, the only primes $p$ that are congruent 1 modulo both 4 and 3, are of the form $12m+1$, hence $<Q_1(x,y)>_p \cap <Q_2(x,y)>_p=\{12m+1| 12m+1 \in\mathbb{P}\}$. This gets us closer, but this is not what you asked for.
Now let's find some integers for which we are guaranteed never to get any of this primes on one of the quadratic forms!!
Let's fix $y$ to $1$. So now we have,
$1) \; \;Q_1(x,1)=Q_1(x)=x^2+3$ 
$2) \; \; Q_2(x,1)=Q_1(x)=x^2+1$
Notice that 
$$ Q_{1}(x)\mod{12} \equiv \{3,4,7,0\}$$
Hence,  primes of the form $p \mod12 \equiv1$ never appear for any $x$ in $Q_1(x)$. Also, note that both $Q_1(x),Q_2(x)$ irreducible over $\mathbb{Q}$. Since we meet the second condition stated earlier about the primes generated by $Q_1$ and $Q_2$ we are done.
And Tada!!! That is all. :)
A: Let $f(y)=y^2-2$, and let $g(x)=x^3+5$, then this works, because if they had a solution in common you'd have a solution for $y^2=x^3+7$, but this is known not to have any integer solutions. See Silverman's Arithmetic of Elliptic curves page 296.
