Is it true that for every polynomial $P(x)$, there exist an integer $n$ so that $Q(x)=P(x)+n$ is irreducible

Question

Is it true that for every polynomial $P(x)$, there exist an integer $n$ so that $Q(x)=P(x)+n$ is irreducible over the rational numbers? If not, what are the conditions of $P(x)$ so that there exist an integer $n$ so that $Q(x)=P(x)+n$ is irreducible? If $n$ is not an integer but a rational number, will the statement still correct?

I tried to use the Eisenstein criterion, but I couldn't make any progress from that.

Answer 1

We can choose the constant term to be coprime to all other coefficients, so the polynomial is primitive. Then by Gauss's lemma we know that irreducibility in $\mathbb{Q}$ is equivalent to irreducibility in $\mathbb{Z}$.

Now suppose that $f(x)$ is reducible in $\mathbb{Q}$. Then, it's reducible in $\mathbb{Z}$, so $f(x) = g(x) h(x)$ for some non-constant polynomials $g, h \in \mathbb{Z}[x]$.

If the constant term of $f$ is prime, then since $f(0)=g(0) h(0)$, we have that the constant term of $g$ or the constant term of $h$ is $\pm$ the constant term of $f$.

Then, if we choose the constant term of $f$ to be extremely large, then either $g$ or $h$ must also have that large constant term. Thus one of the other coefficients of $f$ must also be extremely large.

Can you finish the proof from here?