Compare $P(-1,y)$ and $P(1,y)$, $f(-y)=\frac{f(-1)}{f(1)}f(y),y\ne\pm1$ Swap $-y,y$, we get $f(-1)^2=f(1)^2$. So $f$ is either even of odd. If $f$ is even, compare $P(x,y)$ and $P(x,-y)$, $f(x^2+y)=f(x^2-y)$. It is easy to show $f$ is constant. So the first solution is $f(x)=c, c\ne0$. Now assume $f$ is odd. 1) $f(x+1)=f(x)+1$ and $f(2x)=2f(x)$. $P(x,-1):f(x^2-1)+2=f(x^2+1),x^2\ne0,1$. $P(1,x):f(1+x)=\frac{f(x)}{f(1)}+f(2)-1,x\ne0,-1$ implies $f(x+2)=\frac{f(x)}{f(1)^2}+(f(2)-1)(1+\frac1{f(1)})$. Compare the above two, $f(f(2)-1)(1+\frac1{f(1)})=2$ and $f(1)^2=1$ Hence, $f(1)=1,f(2)=2$, $f(x+1)=f(x)+1,x\ne0,-1$. $P(x,2):f(2x)=2f(x),x>0$. Since $f$ is odd, $f(2x)=2f(x)$. 2) $f(x)=x$. $P(\sqrt{x},y)+P(\sqrt{x},-y):f(x+y)+f(x-y)=2f(x)=f(2x),x>0,x\ne\pm y$. Easily we can show Cauchy equation holds. Then the f.e. also implies $f(xy)=f(x)f(y),x>0,x,y>0$ So $f(x)=x$ is the other solution. The solutions are $f(x)=c,c\ne0$ and $f(x)=x$.

Nice solution!! ((At least this one is shorter than mine

For f-$odd$: $P(x,-1):f(x^2-1)+2=f(x^2+1)$, i.e. $f(x+2)=f(x)+2$ $P(x,3): f(3x)=3f(x)$ , so $f(3)=f(1)+2$ and $f(3)=3f(1)$, we conclude $f(1)=1$ Thus, $f(x+1)=f(x)+1$, then $P(x,x):f(x^2+x)=f(x^2)+\frac{f(x^2)}{f(x)}$ Then $P(x,x+1)$ gives $f(x^2)=f(x)^2$ Then substracting $P(x,y)$ and $P(x,-y)$ as above we have adiitivity $f(x^2)=f(x)^2$ and additivity implies monotonousity and it's solved in reals $f(x)=f(1)x=x$

Lots of the proposers of ELMO problems are current USA contestans. For whom the contest is specified?

math90 wrote: Lots of the proposers of ELMO problems are current USA contestans. For whom the contest is specified? You can read this Quote: The ELMO is an IMO-style exam prepared by returning MOPpers for the new MOPpers. I think this means that ELMO's proposers are returning MOPers.

Why $f(-1)^2=f(1)^2$ implies f is either even or odd.

hello.. anyone please clarify.

Since $f(-y)=\frac{f(-1)}{f(1)}f(y)$ for $y\notin\{-1,1\}$. If $f(1)=f(-1)$ then $f(-y)=f(y)$. If $f(1)=-f(-1)$ then $f(-y)=-f(y)$.