irreducibility of polynomials made by perturbation from a polynomial Suppose $f(x)\in\mathbb{Z}[x]$ with $\text{deg}f=2n,n\in\mathbb{Z_+}$ and $f_m(x):=f(x)+ mx^n $ for each integer $m\in\mathbb{Z}$. Let us define a number $P_f$:
$P_f:=|\{f_m(x)\in\mathbb{Z}[x]:m\in\mathbb{Z},f_m\ \text{reducible}\}|.$
It seems to be able to prove $P_f<\infty$ for any $f$. For example, if $\text{deg}f=2$, we may use Perron's criterion or just simple factoring calculation. A proof may be constructed based on the fact that an integer has only finitely many factors, for higher degree cases.
Question: How large can $P_f$ be? I guess $P_f\le 1$(I should replace it by $P_f\le 2$; see an answer below.) if $\text{deg}(f)\ge 10$. Is there a counterexample for this conjecture? If there is a counterexample, is there any positive integer $n$ such that $P_f\le 2$ if $\text{deg}(f)\ge n$? What if we consider polynomials with only positive coefficients or only palindromic(i.e., self-reciprocal) polynomials?
 A: Fuller answer:
Let $f(x) = x^{2n} + p^nc$. Then $f_{-(p^n + c)}(p) = p^{2n} - p^n(p^n + c) + p^nc = 0$. So if $f(x) = x^{2n} + a^n$, then for each distinct factor $b$ of $a$, there exists some value of $m$ such that $f_m(b) = 0$, and so $f_m$ is reducible. So if $a$ has lots of distinct factors, then there are lots of values of $m$ for which $f_m$ is reducible. (Potentially, some factors of $a$ could give the same value of $m$, but at most $2n$, since $f_m$ has at most $2n$ roots). So $P_f$ can be arbitrarily large for any fixed $n$.
The question becomes harder if we assume that $f$ is monic and palindromic, so only $\pm 1$ can be roots of $f_m$ for some $m$ (since $f_m(a) = 1 \mod a$ for any $a, m$). Here it might be true that $P_f \leq 2$.
Note that $\pm 1$ are always roots of $f_m$ for some $m$: assume, wlog, that the coefficient of $x^n$ in $f$ is $0$; then $f_{-f(1)} = f(1) - f(1)1^n = 0$, and similiarly $f_{(-1)^{n+1}f(-1)} = f(-1) + (-1)^{n+1}f(-1)(-1)^n = 0$. So $P_f \geq 1$ for all $f$, and $P_f \geq 2$ unless $f(1) = (-1)^nf(-1)$.
A: There are examples with $P_f = 2$ for all choices of $n$: let $f(x) = x^{2n} - 2x^n + 1$. Then $f_0(x) = (x^n - 1)^2$ and $f_4(x) = (x^n + 1)^2$.
