$f(x)=0$, $f(x)=-x$

EaZ_Shadow wrote: $f(x)=0$, $f(x)=-x$ no

Strange F.E. indeed, denote $P(x,y)$ the assertion of the F.E. $P(0,0)$ gives $f(0)=0$ or $f(0)=-1$. Case 1: $f(0)=0$. Here $P(x,0)$ gives $f(x^2)=f(x)$ meaning that by swapping sign one gets $f$ even, now by $P(0,x)$ one gets that $f(f(x))=-f(x)$, and from $P(x,-x)$ one finishes by getting $f(x)^2=0$ which means $f$ is zero everywhere. Case 2: $f(0)=-1$. $P(0,x)$ gives $f(f(x)+1)=-1$, also $P(x,0)$ gives $f(x^2)=-1$ which means $f(u)=-1$ for all $u \ge 0$. Now pick $x>0$ and $y<0$ but so that $x+y>0$ which would give $-1+(-1)f(y)=0$ which means $f(y)=-1$ for all negatives $y$ as well. Therefore only $f(x)=0,-1$ for all reals $x$ work thus we are done .

@above, why does $P(0, x)$ give $f(f(x))=f(x)$?

$P(0, 0)\implies f(0)=0$ or $f(0)=-1$. $\underline{f(0)=-1}: \ P(x, 0)\implies f(0)+f(x)f(0)=f(x^2)-f(x)\rightarrow f(x^2)=-1\implies\boxed{f(x)=-1, x\geq 0}$ $P(x, -x), x\geq 0\implies f(\underbrace{f(0)-f(x)}_{0})+\underbrace{f(x)}_{-1}f(-x)=\underbrace{f(x^2)-f(0)}_{0}\implies f(-x)=-1\implies\boxed{f(x)=-1 \ \forall \ x}$ $\underline{f(0)=0}: \ P(x, 0)\implies \boxed{f(x)=f(x^2)}\implies\boxed{f(x)=f(-x)}, \ P(0, x)\implies\boxed{f(f(x))=-f(x)}$ $P(f(x), -f(x))\implies f(\underbrace{-f(f(x))}_{f(x)})+\underbrace{f(f(x))}_{-f(x)}f(-f(x))=\underbrace{f(f(x)^2)}_{f(f(x))}\implies f(f(x))-f(x)f(-f(x))=f(f(x))\implies\boxed{f(x)f(-f(x))=0}$ $f(-f(x))=0, \ P(x, -x)\implies \underbrace{f(-f(x))}_{0}+\underbrace{f(x)f(-x)}_{f(x)^2}=\underbrace{f(x^2)}_{f(x)}\implies\boxed{f(x)(1-f(x))=0}$ But $f(f(x))=-f(x)$ and the left side is either $1$ or $0$, so $f(x)$ must be $0\implies \boxed{f(x)=0 \ \forall \ x}$ So the answer is $f(x)=0 \ \forall \ x$ or $f(x)=1 \ \forall \ x$ $_{\blacksquare}$

This problem was proposed by me (Gabriel from Costa Rica). I came up with it back in 2022 while training for what would be my first IMO. Here is the original outline of the proof:

Answer: $f(x)=0,-1 \forall x \in \mathbb{R}$ It's easy to see that these work. Solution: Let $P(x,y)$-denote the given assertion. By taking $P(0,0) \implies f(0)+f(0)^2=0 \implies f(0)(f(0)+1)=0 \implies f(0)=0 \vee f(0)=-1$ Case 1. $f(0)=0$ $P(x,0) \implies \boxed{f(x^2)=f(x)} ...(*) , x \rightarrow -x \implies f(x)=f(x^2)=f(-x) \implies f(x)=f(-x) \implies \boxed{f-\text{even}}$ $P(0,x) \stackrel{(*)}{\implies} \boxed{f(f(x))=-f(x)} ...(2)$ Finally $P(x,-x) \implies f(-f(x))+f(x)f(-x)=f(x) \stackrel{f-\text{even}}{\implies} f(f(x))+f(x)^2=f(x) \stackrel{(2)}{\implies} -f(x)+f(x)^2=f(x) \implies f(x)^2-2f(x)=0 \implies f(x)(f(x)-2)=0 \implies f(x)=0 \vee f(x)=2 \implies f \equiv 0 \vee f \equiv 2.$ Note that $f \equiv 2$ doesn't work but $f \equiv 0 $ does. Case 2. $f(0)=-1$ Claim: $f(x)=-1 \forall x \in \mathbb{R^+}$ Proof: $P(x,0) \implies f(x^2)=-1$ by taking $y=x^2$ hence $y \in \mathbb{R^+}$ we get: $f(y)=-1 \forall y \in \mathbb{R^+} \implies f(x)=-1 \forall x \in \mathbb{R^+}$ $\square$. Claim: $f(x)=-1 \forall x \in \mathbb{R^-}$ Proof: Pick a $y \in \mathbb{R^+}$ so by taking $P(y,-y)$ and using the above claim we get: $f(-2)-f(-y)=0 \implies f(-y)=f(-2) \forall y \in \mathbb{R^-}$. Let $f(-2)=e$ , $e-$constant $\iff f(-y)=e$ so by taking a $z$ s.t $z=-y$ hence $z \in \mathbb{R^+} \implies f(z)=e \forall z \in \mathbb{R+}$ Let $(a,b) \in \mathbb{R^-}$ $P(a,b) \implies (e+1)^2=0 \implies e=-1$ therefore $f(x)=-1 \forall x \in \mathbb{R-}$ $\square$ Now since $f(x)=-1 \forall x \in \mathbb{R^+}$ and $f(x)=-1 \forall x \in \mathbb{R^-}$ we get $f(x)=-1 \forall x \in \mathbb{R}$ $\blacksquare$