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

Let $p\ge 5$ be a prime.

As an easy warm-up 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 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 ($\circ$).

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

Let $p\ge 5$ be a prime.

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. The easiest way to see this is by comparing the images $A_i:=\mathrm{Im}(\phi_i)$: $$ A_1+A_2 \subseteq A_3,\ A_3-A_1\subseteq A_2,\ A_3-A_2\subseteq A_1, $$ whence $|A_1|=|A_2|=|A_3|=1$.

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 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 two different ways, is that there are at least $p-1$ pairs $(x,y)$ satisfying ($\circ$).

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

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 ($\circ$).

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

Let $p\ge 5$ be a prime.

As an easy warm-up 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 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 ($\circ$).