Solution.

u don't know that f is surjective

\above there exist $a$ such that $f(a)=0$ because for $x=f(0)$ and $y=0$ we have $f(2f(0))=0$

oopa im dum

@above no we don't, $P(f(0),0)$ gives $f(2f(0))+f(0)=f(0)$.

Let $P(x,y)$ denote the given assertion. $P(f(0),0)$ implies $f(2f(0))=0.$ And $P(x,2f(0))$ implies $f(x)=x/2$ and this works.

Answer : $ f(x) = \frac {x} {2} $ . Solution : Let $ P(x,y) $ denote the given assertion. $ P(f(x),x) \Rightarrow f(2f(x)) = x - f(0) \Rightarrow f $ is surjective (1). By (1) : $ \exists \alpha : f(\alpha) = 0 $ . $ P(x,\alpha) \Rightarrow f(x) = \frac {x} {2} $ Which is clearly true. As a result, our proof is complete.

The only solution is $f(x) = \frac x2$, which works. Now we show it's the only one. Let $P(x,y)$ denote the given assertion. $P(f(y), y): f(2f(y)) = f(y) - f(0)$. Setting $y = 0$ here gives that $f(2f(0)) = 0$. Setting $y = 2f(0)$ now gives that $f(0) = -f(0)$, so $f(0) = 0$. $P(x,0): 2f(x) = x$, so $f(x) = \frac x2$.