greatest common divisor of polynomials with two variables Let $F$ be an algebraic closed field of char(F)=0 and let $(a,b)\in F^2$ be a non-zero vector. Suppose I have two polynomials  $f,g\in F[x]$ with $\deg f\neq \deg g$ and $1\leq \deg f, \deg g$. I want to show that 
\begin{equation}
\gcd(f(x)-f(y)-a, g(x)-g(y)-b)=1.
\end{equation}
For instance, by Eisenstein criterion, it is easy to verify this for an arbitrary non-zero vector $(a,b)$ and $f(x)=x^n$ and $g(x)=x^m$ when $n\neq m$. I have calculated several other examples which again verify the claim but still I am not sure whether it is true in general. 
I would be thankful if you share with me your ideas. 
 A: Not a complete argument but here are some ideas. Suppose $f$ has degree $d$ and let $c_1,\ldots,c_{d-1}$ be the zeros of $f'(x)$. Consider the curve $X: f(x)-f(y)-a$ in the projective plane. It can only be singular at points of the form $(c_i,c_j)$ with $a=f(c_i)-f(c_j)$. So if $a$ is not one of those $(d-1)(d-2)$ values then $X$ is smooth, hence irreducible and the gcd is $1$ for every $g, \deg g < \deg f$. Even if $a$ is one of the "bad" values, as long as the bad value does not repeat for many pairs $(c_i,c_j)$ we can still conclude that $X$ is irreducible. Namely, if $X$ is the union of two curves of degrees $m,d-m$ it has at least $m(d-m)>d-1$ singularities where the two curves meet. (One needs to be careful with intersections of high multiplicity here, I am ignoring these subtleties). I think it should be easy to rule out $m=1$ so you might even improve the number of coincidental $(c_i,c_j)$ to $2(d-2)$. Is it even possible to construct an $f$ such that $f(c_i)-f(c_j)$ takes the same value for at least $2(d-2)$ pairs $(c_i,c_j)$ of roots of $f'$? 
