Let $h(x)=f(x)-f(0)$. We have $$h(x+y)=f(x+y)-f(0)=f(x+y)-f(y)+f(y)-f(0)=f(x)-f(0)+f(y)-f(0)=h(x)+h(y).$$This is the Cauchy equation. Since $2h(x)$ is always an integer, we conclude that $h(x)\equiv 0$. Hence $f(x)=n/2\ (n\in\mathbb{Z})$.

Andrew_Sakura wrote: Let $h(x)=f(x)-f(0)$. We have $$h(x+y)=f(x+y)-f(0)=f(x+y)-f(y)+f(y)-f(0)=f(x)-f(0)+f(y)-f(0)=h(x)+h(y).$$This is the Cauchy equation. Since $2h(x)$ is always an integer, we conclude that $h(x)\equiv 0$. Hence $f(x)=n/2\ (n\in\mathbb{Z})$. shouldn't it be $f(x)\equiv 0$? What's more,Cauchy Equation can be only solved when the function is continuous or monotonic.

MathLover_ZJ wrote: Andrew_Sakura wrote: Let $h(x)=f(x)-f(0)$. We have $$h(x+y)=f(x+y)-f(0)=f(x+y)-f(y)+f(y)-f(0)=f(x)-f(0)+f(y)-f(0)=h(x)+h(y).$$This is the Cauchy equation. Since $2h(x)$ is always an integer, we conclude that $h(x)\equiv 0$. Hence $f(x)=n/2\ (n\in\mathbb{Z})$. shouldn't it be $f(x)\equiv 0$? What's more,Cauchy Equation can be only solved when the function is continuous or monotonic. $f(x)\equiv n/2$ and $g(x)\equiv n$. The solutions of Cauchy equation would be either linear, or dense in $\mathbb{R}^2$.

Andrew_Sakura wrote: Let $h(x)=f(x)-f(0)$. We have $$h(x+y)=f(x+y)-f(0)=f(x+y)-f(y)+f(y)-f(0)=f(x)-f(0)+f(y)-f(0)=h(x)+h(y).$$This is the Cauchy equation. Since $2h(x)$ is always an integer, we conclude that $h(x)\equiv 0$. Hence $f(x)=n/2\ (n\in\mathbb{Z})$. How did you conclude $f$ is a constant?

PNT wrote: How did you conclude $f$ is a constant? Because $f(x)-f(0)=h(x)=0\implies f(x)=f(0)=\text{constant}$

ZETA_in_olympiad wrote: PNT wrote: How did you conclude $f$ is a constant? Because $f(x)-f(0)=h(x)=0\implies f(x)=f(0)=\text{constant}$ But why $h(x)=0$ ?

PNT wrote: But why $h(x)=0$ ? As shown, function $h$ is additive. Because $2h(x)\in \bf Z$ for all $x$, the graph of $h$ is not dense in the plane. So it must be linear, i.e. $h(x)=k\cdot x$. Again since $2h(x)= 2k\cdot x \in \bf Z$ for all $x$, we must have $k=0$. Hence $h(x)=0$.

By comparing $P(x+y,0)$ and $P(x,y)$ we get $f(x+y)+f(0)=f(x)+f(y)$; when we sub $h(x)=f(x)-f(0)$ we get that $h$ is additive. Also, $2f(0)\in\mathbb Z$ by $P(0,0)$ and $P(x,x)$ is equivalent to $2h(x)=\lfloor g(x+y)\rfloor-2f(0)$, so $2h(x)\in\mathbb Z$. Therefore, $h$ "omits some disk" (consider one contained between $y=\frac12$ and $y=0$) and $h$ must be linear. In particular, $h(0)=0$, so let $h(x)=ax$. We have $2ax\in\mathbb Z$ for all $x$, so $a=0$ necessarily (otherwise $x=\frac1{4a}$ yields a contradiction). Then $f$ is constant, so $\boxed{f(x)=\frac n2}$ for some $n\in\mathbb Z$. This works, with $g(x+y)=n$.