Let $f(x,y) \in \mathbb{C}[x,y]$ and let $a,b \in \mathbb{C}$, $a \neq b$. Assume that $f(a,y)=f(b,y)=1$.
Is it true that $f(x,y)=(x-a)(x-b)g(x,y)+1$, for some $g(x,y) \in \mathbb{C}[x,y]$?
It is clear that if $f(x,y)=(x-a)(x-b)g(x,y)+1$, for some $g(x,y) \in \mathbb{C}[x,y]$, then $f(a,y)=f(b,y)=1$, but what about the opposite direction?
Remark: We can consider $h(x,y):=f(x,y)-1$, and so $h(a,y)=h(b,y)=0$.
Any hints are welcome!
The converse also holds.
Let $f(x,y) =\sum_{i,j}a_{i,j}x^i y^j = (x-a)Q(x,y) + P(y) $, for some polys P and Q. Since $f(a,y) = 1 \forall y $, $P(y) = 1, \forall y$ follows.
So $f(x,y) = (x-a)Q(x,y) + 1$
Playing the same game with $b$ and $Q(x,y)$ yields that $$f(x,y) = (x-a)((x-b)H(x,y) + G(y))+1= (x-a)(x-b)H(x,y) + (x-a)G(y)+1.$$ Since $f(b,y)=1, \forall y$ it follows that $(b-a)G(y)=0$ $\forall y$ so $G=0$ since $b \neq a$.
So $f(x,y) = (x-a)(x-b)H(x,y) + 1 $ as desired.