https://aops.com/community/p11377651 rmtf1111 wrote: Let $c=f(0)$. For any reals $x,y$ we have that $P(x,y)-P(-x,y) \implies f(x+y)=f(y-x)$ or $f(x+y)+f(y-x)=2y$. Plug-in $y:=x\implies f(x)=x-c$ or $f(x)=x$. The original assertion implies that $f(z)\ge x$ for any $z\ge f(x)$, thus for sufficiently large $x$ we have that $f(x)=x-c$ only. Let $z$ be an arbitrarily small number, now it is easy to see that $P(z,y-z)$ finishes the problem.

$p(x,y);p(-x,y)$

lazizbek42 wrote: $p(x,y);p(-x,y)$ This is the same as post #2.