https://aops.com/community/p11377651 rmtf1111 wrote: Let $c=f(0)$. Plug-in $z:=x+y$ and $w:=0$, then $f(x)+f(y)=f(x+y)+c$ and blah blah cauchy

$P(x,y,x+y,0)$

lazizbek42 wrote: $P(x,y,x+y,0)$ This is the same as post #2.

Let $x,y,z,w$ be such that $x+y = z+w$. Then we have $f(x)+f(y)=f(z)+f(w)$. Here, $(x,y,z,w)= (x,y,x+y,0)$ gives us that $$f(x)+f(y)=f(x+y)+f(0),$$so $g(x)= f(x)-f(0)$ is an additive function on $\mathbb Q$ and then $f(x) = ax+b$ for some rationals $a,b$. Plugging this to the main assumption of the problem we see all rationals $a,b$ work.