$n,m\ge 1$, we have $$\deg\gcd\(f^n-1,g^m-1\)\le \(11(d_f+d_g)^2\)^{\min(d_f,d_g)}.$$

The proof, except the explicit bound for the degree, is given in [@AR Theorem 1]. To see the degree bound, we just apply Lemma \[lem:BS\].

Our curve is given in parametric form $\{\(f(t),g(t)\)~:~ t\in\C\}$ and we need to find the degree of the implicit form $H$ such that $H(f(t),g(t))=0$, $t\in\C$. This is obtained using resultants, that is $$H={{\operatorname{Res}}}_T\(f(T)-X_1,g(T)-X_2\),$$ which is a polynomial of degree $\deg g$ in $X_1$ and $\deg f$ in $X_2$. Thus, the total degree of $H$ is at most $\deg f+\deg g$.

Let $\widetilde{H}$ be an absolutely irreducible factor of $H$ and assume that $\widetilde{H}(f(t),g(t))=0$ for infinitely many $t\in\C$. As $\widetilde{H}(f(T),g(T))$ is a univariate polynomial , we must have the identity $\widetilde{H}(f(T)),g(T))=0$. Then, by Lemma \[lem:BS\] applied with the curve defined by the polynomial $\widetilde{H}$, we obtain that $\widetilde{H}$ is of the form $X_1^{n_1}X_2^{n_2}=\omega$, for some root of unity $\omega$ and integers $n_1,n_2$ not both zero. This implies that $f,g$ are multiplicatively dependent, which contradicts the hypothesis. Thus, there is no such absolutely irreducible divisor of $H$.

Therefore $\gcd\(f^n-1,g^m-1\)$ has at most $11(d_f+d_g)^2$ distinct zeros. As in the proof of [@AR Theorem 1], we see that the multiplicity of each zero 