The question: Let $g, f_1, f_2$ be homogeneous polynomials over the complex numbers in the variables $x_0, x_1, x_2$, assume $g$ is irreducible and of degree $m$ and $f_1, f_2$ are of degree $n > m$. We will look at the algebraic curves associated with these polynomials in the complex projective plane. If $V(f_1)$ and $V(f_2)$ intersect in precisely $n^2$ points and $V(g)$ intersects both in $mn$ or these points, then find a linear combination $f$ of $f_1$ and $f_2$ such that $g$ and $f$ intersect in more than $mn$ points.
What I’ve tried: clearly, $f$ will usually also be of degree $n$, and surely of degree at most $n$, and with $g$ being of degree $m$ it can’t be that $f$ and $g$ intersect in more than $mn$ points unless they share a common component (using Bézout’s theorem). Since $g$ is irreducible, this means that $g$ must be that “common component” (or better, $V(g)$), ie, $g$ divides $f$. We hence have to find complex numbers $a$ and $b$ and a polynomial $h$ of degree $n-m$ such that $$af_1 + bf_2 = gh.$$ This is unfortunately where I get stuck. Any help is appreciated!
I believe I have answered my own question. (Feedback still welcome!)
Pick a point $X$ on $V(g)$ different from the $mn$ it shares with $f_1$ and $f_2$. Since we are working with projective invariants, we can assume $X = E_0 = [1;0;0]$ after applying a projective transformation. By Bézout’s theorem, $E_0$ is not a point on $V(f_1)$ or $V(f_2)$, hence if we order the terms in $x_0$, we may rewrite both polynomials as $$f_i(x_0,x_1,x_2) = \alpha_{0i} x_0^n + \alpha_{1i}(x_1, x_2) x_0^{n-1} + \ldots$$ for some non zero constants $\alpha_{01}, \alpha_{02}$. Then $f = \alpha_{02} f_1 - \alpha_{01} f_2$ is the linear combination which we were looking for. Indeed, $V(f)$ contains the $mn$ points in the intersection and also $E_0$.