37th Iranian Mathematical Olympiad

The answers are $f(x)=x$, $f(x)=-x$ and $f(x)=0$.

First, let's prove that $f(x)$ is injective at point $0$. Assume there exist two distinct real numbers $t_1$ and $t_2$ such that $f(t_1)=f(t_2)=0$. Comparing $P(-f(b)-f(0),b,t_1)$ and $P(-f(b)-f(0),b,t_2)$ gives us $$(f(b)+f(0))b{t}_1=(f(b)+f(0))b{t}_2.$$ So if $f(x)$ is not injective at point $0$, we have $$\Rightarrow \forall b\ne 0:f(b)=-f(0).$$ Which gives us $f(x)=0$ as an answer.

So if $f(x)$ is a non-constant function, then it must be injective at point $0$.

Now, $P(-f(0)-f(f(0)),0,0)$ gives us $f(-f(0)-f(f(0)))=0$ and we have a real number $t$ such that $f(t)=0$. $P(-f(f(0)),t,0)$ gives us $$f(-f(f(0)))=f(-f(0)-f(f(0)))=0.$$ And according to the injectivity at point $0$, it follows that $f(0)=0$.

Now, $P(-f(f(b)),b,0)$ and $P(-f(f(c)),0,c)$ give us $$f(b)f(f(b){)}^2=b^{2}f(b).$$ If $b\ne 0$ then $f(b)\ne 0$ therefore $$f(f(b))=\pm b\forall b\ne 0.$$ And since $f(f(0))=f(0)=0$, we have $$f(f(b))=\pm b\forall b\in \mathbb{R}.$$ If there exists a real number $c\ne 0$ such that $f(f(c))=-c$, $P(c,0,c)$ gives us $$f(c)(f(c{)}^2+c^2)=0$$ which is a contradiction.

So $$\forall c\in \mathbb{R}:f(f(c))=c$$ Now $P(-a,0,a)$ gives us $f(-a{)}^2+a^{2}f(a)=0$ and $P(a,0,-a)$ gives us $f(a{)}^3+a^{2}f(-a)=0$ and they lead to $f(a)=\pm a$.

If there exist non-zero real numbers $b,c$ such that $f(b)=b$ and $f(c)=-c$, $P(-b-c,b,c)$ leads to contradiction. So $f(x)=x$ and $f(x)=-x$ are the only solutions.

$f(x)=0$ is also a solution.

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