Let $Q(x,y)$ be assertion $f(x+y)=f(x)+f(y)$, and $P(x)=ax^2+bx+c$ such that $a\neq 0$. $Q(P(x), P(x+1)-P(x))\implies f(P(x+1))=f(P(x))+f(P(x+1)-P(x))$ $f(x+1)=f(x)+f(2ax+a+b)\implies f(1)=f(2ax+a+b)$ Since $a\neq 0$, we can plug in $x\mapsto \frac{x-a-b}{2a}$ and get that $f(x)=f(1)$ so $f$ is constant. $Q(0,0)\implies f(0)=f(0)+f(0)\implies f(0)=0$. Thus, $f(x)=0\quad\forall x\in\mathbb{R}$.

Quidditch wrote: $(Expand)$ Let $P$ be a given polynomial. Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that $$ \exists M: \, f(x+y)=f(x)+f(y), \, x \in R, \, y \geq M \text{ and } f(P(x))=f(x)\text{ for all } x \in R.$$ Claim 1: $f(x+y)=f(x)+f(y) \forall x,y \in R, (1)$ Proof: For any $x,y \in R$, choose $z: \left\{\begin{matrix}z > M & \\z+x > M & \\ \end{matrix}\right.$ Then $\left\{\begin{matrix}f(z+x+y)=f(z)+f(x+y) & \\f(z+x+y)=f(z+x)+f(y)=f(z)+f(x)+f(y) & \\ \end{matrix}\right. \Rightarrow f(x+y)=f(x)+f(y) \forall x,y \in R$ (Q.E.D) Let $x=y=0$ in $(1)$, we get $f(0)=0$ Let $y=-x$ in $(1)$, we get: $f(x)+f(-x)=0$ or $f$ is odd If $P(x)$ satisfied then $-P(x)$ also satisfied, so we can assume the coefficient of the second highest degree bigger than $0$ So the leading coefficient of $P(x+1)-P(x)$ bigger than $0$ Do the same as above, we get $f(P(x+1)-P(x))=f(1)$, which means $\exists c: \, f(x)=f(1), \forall x > c$ Now for any $x \in R,$ choose $y_0: \left\{\begin{matrix}y_0 > c & \\y_0+x > c & \\ \end{matrix}\right.$ Then let $y=y_0$ in $(1)$, we get $f(x)=0, \, \forall x \in R$