Find all $P,Q \in \mathbb{Q}[X]$ such that $P+aQ$ is irreducible for every rational number $a$.
If $\deg P, \deg Q \leq 1$, then every $P,Q \in \mathbb{Q}[X]$ satisfy the hypothesis. Also, $P$ is irreducible and if $(P,Q)=R$ and $\deg R\geq 1$, then $P+aQ$ is reducible, unless $P=cQ$ for some $c \in \mathbb{Q}$. My guess is these are the only solutions, but I can't prove this.
Suppose that neither $P(x)$ nor $Q(x)$ are identically $0$.
Assume we had a counterexample.
There are, then, infinitely many $x_0$ with $Q(x_0)\neq 0$ and of course $$F_{x_0}(x)=P(x)-\frac {P(x_0)}{Q(x_0)}Q(x)$$ has a root at $x_0$. Thus it must be the case that each of the $F_{x_0}(x)$ has degree $≤1$. But if degree $P(x)$ is greater than $1$ then $P(x)-aQ(x)$ has degree $≤1$ for (at most) one choice of $a$. Call that choice $A$. Thus we see $$\frac {P(x_0)}{Q(x_0)}=A\implies P(x_0)=Q(x_0)\times A$$ for infinitely many $x_0$. Whence we must have $P(x)=Q(x)\times A$ and we are done.