Can we find $\lambda \in \mathbb{C}$ such that $\deg(\gcd(f(t-\lambda),g(t-\lambda)))=1$?

Question

Let $f=f(t),g=g(t) \in \mathbb{C}[t]$ with $\deg(f) \geq 2$ and $\deg(g) \geq 2$.

Can we find $\lambda \in \mathbb{C}$ such that $\gcd(f(t-\lambda),g(t-\lambda))$ is of degree $1$?

Please notice that a somewhat similar question to the above question has a positive answer, namely, given $f$ and $g$ of degrees $\geq 2$, there exists $a,b \in \mathbb{C}$ such that $\gcd(f(t)-a,g(t)-b)$ is of degree $1$.

Thank you very much!

Answer 1

Not really, because if $u=\operatorname{gcd}(f,g)$, then $\operatorname{gcd}(f(t-\lambda),g(t-\lambda))=u(t-\lambda)$.