Let $f(t)=0$ for some $t$ $y=t:f(-f(x))=1-x-t$ $f(-f(-f(x)-f(y)))=f(y+x-1)=1+f(x)+f(y)-t$ $y=0:f(x-1)=f(x)+f(0)-t+1$ So $ f(y+x)=f(x+1)+f(y)-t+1=f(y)+f(x)-f(0) \to f(x)=ax+f(0)$ $f(-f(x)-f(y))=f( -a(x+y)-2f(0))=-a^2(x+y)-2af(0)+f(0)=1-x-y \to a^2=1, f(0)-2af(0)=1 \to f(x)=x-1$ Or more faster solution: $f(-f(x)-f(0))=1-x \to f(x)$ is bijection $f(-f(x+y)-f(0))=1-x-y=f(-f(x)-f(y)) \to f(x+y)+f(0)=f(x)+f(y) \to f(x)=ax+f(0)$ $1-x=f(-f(x)-f(0))=f(-ax-2f(0))=-a^2x-2af(0))+f(0) \to a^2=1, (2a-1)f(0)=-1 \to f(x)=x-1$

What Motivation was behind the last line?

This is Singapore MO 2019 P2

Compare $P(x, y+1)$ and $P(x+1, y)$. RHS for both is $-x-y$ so $f(-f(x)-f(y+1))=f(-f(x+1)-f(y))$ using injectivity $f(x+1)-f(x)=f(y+1)-f(y)$, since that $f$ is linear