$ \newcommand{\F}{{\mathbb F}}$ $ \newcommand{\R}{{\mathbb R}}$ $ \renewcommand{\phi}{\varphi}$

Let $ p\ge 5$ be a prime.

As an easy exercise, if the functions $ \phi_1,\phi_2,\phi_3\colon\F_p\to\R$ satisfy $ \phi_1(x)+\phi_2(y)=\phi_3(x+y)$ for all pairs $ (x,y)\in\F_p^2$ , then they are constant functions.

Given that $ \phi_1,\phi_2,\phi_3\colon\F_p\to\R$ are non-constant, what is the largest possible number of pairs $ (x,y)\in\F_p^2$ satisfying $ $ \phi_1(x)+\phi_2(y)=\phi_3(x+y)? \tag{$ \ast$ } $ $ Equivalently, what is the smallest possible number of pairs $ (x,y)\in\F_p^2$ , for all possible choices of the (non-constant) functions $ \phi_1,\phi_2,\phi_3$ , such that ($ \ast$ ) fails to hold?

If $ u,v,w\in\F_p$ are pairwise distinct and $ w\ne u+v$ then, letting $ $ \phi_1=1_{\{u\}},\ \phi_2=1_{\{v\}},\ \phi_3=1_{\{w\}} $ $ (the indicator functions of the corresponding singletons), we have $ 3p-5$ pairs $ (x,y)$ violating ($ \ast$ ); is this the worst case?

Is it true that for any non-constant functions $ \phi_1,\phi_2,\phi_3\colon\F_p\to\R$ , there are at least $ 3p-5$ pairs $ (x,y)\in\F_p^2$ with $ $ \phi_1(x)+\phi_2(y) \ne\phi_3(x+y)? \tag{$ \circ$ }$ $

What I can show, in a rather indirect way, is that there are at least $ p-1$ pairs $ (x,y)$ satisfying ($ \circ$ ).