Irreducible polynomials with one large coefficient

Question

Is it true that for every monic polynomial $p(x) \in \mathbb{Z}[x]$, $p(0)\neq 0$, of degree $n>0$ there exists a real number $M>0$ such that for every $|m|>M$ and for every $k$ odd integer ($0<k<n$) , $p(x)+mx^k$ is irreducible over the integers ? If $k$ is even there are counter-examples like, for $k=2$, $$(x^2-nx+1)(x^2+nx+1)=x^4-n^2x^2+1.$$

Answer 1

The polynomial $p(x)=(x^2+1)^3$ is a counterexample. This is because $$ p(x)-n^3x^3=(x^2+1-nx)[(x^2+1)^2+nx(x^2+1)+n^2x^2] $$ is not irreducible as a difference of two cubes, and we can make $n^3$ as large as we wish.

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A very similar variant is the family $$ (x^2+ax+1)(x^4-ax^3+a^2x^2-ax+1)=x^6+x^4+(a^3-2a)x^3+x^2+1, $$ where only the cubic term depends on the parameter $a$. By appropriate choices of $a$ the coefficient of that cubic term can be made arbitrarily large.

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OTOH it is clear that all cubic polynomials $p(x)$ have this property. A monic cubic polynomial in $\Bbb{Z}[x]$ fails to be irreducible only when it has an integer root. The rational root test implies that there are only finitely many choices for that root. So if we fix the quadratic term, the polynomial can be reducible only for finitely many choices of the linear term.

In other words, I think there may be an even more interesting question lurking underneath this. I don't know the best way to reformulate it. But, for example, I failed to produce this kind of a family of sextic polynomials that would factor as a product of two cubics for infinitely many choices of the undetermined coefficient.