I found three solutions but ran out of time to prove them/find more: $f(x)=0$ and $f(x)=\pm\frac{\sqrt{2}}{2}$. Are there any more?

ghu2024 wrote: I found three solutions but ran out of time to prove them/find more: $f(x)=0$ and $f(x)=\pm\frac{\sqrt{2}}{2}$. Are there any more? No, but as you said you should prove it there aren't others.

this is horribly wrong

ghu2024 wrote: This was the solution I submitted because I ran out of time. How many points? The answer is $f\left(x\right)=0$ and $f\left(x\right)=\pm\frac{\sqrt2}{2}$ only, which works. To prove that is all, we interchange $x$, and $y$: ($f(x)+f(y))(1-f(x)f(y))=f(x+y)$ $\left(f\left(y\right)+f\left(x\right)\right)\left(1-f\left(y\right)f\left(x\right)\right)=f\left(y+x\right)$ Note that these two equations are the same, so thus $f\left(x\right)=f\left(y\right)$ equals a constant c. Plugging this in gives: $\left(2c\right)\left(1-c^2\right)=c$, giving us our only other solution, $f\left(x\right)=\pm\frac{\sqrt2}{2}$. This is very wrong in many ways.

Does anyone have a correct solution that they are willing to post?

ghu2024 wrote: Does anyone have a correct solution? I do

I solved the (easy) case when $f(0)$ different from $0$

Let $t$ be an arbitrary real and $x=f(t)$. Then applying the functional equation to $(t, t), (t, 2t), (t, 3t), (2t, 2t)$ we obtain \begin{align*} f(2t) &= 2x(1-x^2) \\ f(3t) &= (2x(1-x^2)+x)(1-2x^2(1-x^2)) \\ &= (3x-2x^3)(1-2x^2+2x^4) \\ f(4t) &= 4x(1-x^2)(1-4x^2(1-x^2)^2)) \\ &= (x+(3x-2x^3)(1-2x^2+2x^4))(1-x(3x-2x^3)(1-2x^2+2x^4)). \end{align*}Now compute \begin{align*} & 4x(1-x^2)(1-4x^2(1-x^2)^2)-(x+(3x-2x^3)(1-2x^2+2x^4))(1-x(3x-2x^3)(1-2x^2+2x^4)) \\ =& 2x^5(x^2-1)(2x^2-1)(2x^2-3)(2x^4-4x^2+3). \end{align*}Therefore the range of $f$ is contained in $\{0, \pm 1, \pm\sqrt{\tfrac{1}{2}}, \pm\sqrt{\tfrac{3}{2}}\}$. We can check that this makes it impossible for $f(2t)=\pm 1$, so we can remove $\pm 1$ from the range. Next, if $f(x)=\pm\sqrt{\tfrac{1}{2}}$ and $f(y)=\pm\sqrt{\tfrac{3}{2}}$, then $f(x+y)$ is not one of the possible numbers in the range, so the range is contained in either $\{0, \pm \sqrt{\tfrac{1}{2}}\}$ or $\{0, \pm \sqrt{\tfrac{3}{2}}\}$. In the latter case, this results in $f(3t)=0$ always, so $f\equiv 0$. Finally, assume the range of $f$ is contained in $\{0, \pm \sqrt{\tfrac{1}{2}}\}$. We can check that $f(x)=\pm\sqrt{\tfrac{1}{2}}, f(x+y)=\mp \sqrt{\tfrac{1}{2}}$ is impossible in the original FE, so assume that the range is contained in $\{0, \sqrt{\tfrac{1}{2}}\}$ (the other case is similar). Then we have $f(x)= \sqrt{\tfrac{1}{2}}\implies f(x+y)= \sqrt{\tfrac{1}{2}}$ in the original FE, so we only get constant solutions. In summary, there are three solutions: $\boxed{f\equiv 0}, \boxed{f\equiv \sqrt{\tfrac{1}{2}}}, \boxed{f\equiv -\sqrt{\tfrac{1}{2}}}$. (Yes, you can solve this problem without using a degree $15$ polynomial. No, I will not do it.)

fakesolve oops

bobjoe123 wrote: Plugging in the equation for (x,x) gives us $2f(x)(1-f(x)^2)=f(x)$ This is wrong. Should be 2x on the RHS

We can get that $f$ can have at most four different values. First $f(0)\neq0$ is easy to show (that there are at most $3$ values). So $f(0)=0$. Now $f(x)=-f(-x)$ or $f(x)=\frac{1}{-f(-x)}$. Considering these two cases for $x$ and using $P(-x,y)$ and $P(y-x,x)$ we can get desired result. ( not expecting more than 2 pts for this result)

Answers Cubic equation has these solutions. $f(0)=\pm\frac{1}{\sqrt 2}$ If $f(0)=+\frac {1}{\sqrt 2}\implies f(x)=\frac {(-1\pm 3)}{(2\sqrt 2)} $ or $f(x)=\frac {(1\pm 3)}{(2\sqrt 2)} $ $f(x)=-f(-x)$ or $f(x)=f(-x)^{-1}$ $f(x)=0$ is also possible. Solution $f(0)\in\{0,-\sqrt{\tfrac 12},\sqrt {\tfrac 12}\}$ If $f(0)=0$ then the equation states $y=0\implies f(x)=f(x) $ If $f(0)=\sqrt {\tfrac 12} $ $f(x)+\sqrt {\tfrac 12}-f(x)^2\sqrt {\tfrac 12}-\frac{f(x)}{2}=\left (f(x)+\sqrt {\tfrac 12}\right)\left(1-f(x)\sqrt {\tfrac 12}\right)=f(x)$ $$\begin {aligned} \implies \sqrt {\tfrac 12}-f(x)^2\sqrt{\tfrac 12}=\frac {f(x)}{2}&\\=f(x)^2+\left (\frac {\sqrt 2}{2}\right)f(x)-1&\\=0\implies f(x)=-\frac {\sqrt 2}{4}\pm\sqrt{\tfrac 12+4}&\\=-\sqrt {\tfrac 12}\pm\frac {3}{\sqrt 2}&\\=\frac {\left (-1\pm 3\right)}{\sqrt 2}\in\{\sqrt 8,\sqrt 2\} \end {aligned} $$$f(x) $ can only take two possible values. The equation works when $x=y=0, c!=0. $ Fix $f(x)\in\{-\sqrt 8,\sqrt {2}\}$ If $c,d!=0$ Then $\left (\sqrt 8+(-\sqrt 8)\right)\left(1-\sqrt {8}^2\right)=-2\sqrt 8 (1-7)=-12\sqrt 8\ne 0,-\sqrt 8$ These are the valid options since we could have $x+y=0,(x+y)!=0$ If $f(x)=\sqrt 2$ $\left (\sqrt 2+\sqrt 2\right)\left(1-\sqrt{2}^2\right)=2\sqrt 2 (1-2)=-2\sqrt 2\notin\{0,\sqrt 2\}$ We can deduce that $f(0)=0$ because the $f(0)!=0$ leads to those contradictions such that $f $ has to be constant on $\mathbb R\setminus\{0\}\implies\text {contradiction} $

This what my partial solution was during the exam.

a1267ab wrote: Let $t$ be an arbitrary real and $x=f(t)$. Then applying the functional equation to $(t, t), (t, 2t), (t, 3t), (2t, 2t)$ we obtain \begin{align*} f(2t) &= 2x(1-x^2) \\ f(3t) &= (2x(1-x^2)+x)(1-2x^2(1-x^2)) \\ &= (3x-2x^3)(1-2x^2+2x^4) \\ f(4t) &= 4x(1-x^2)(1-4x^2(1-x^2)^2)) \\ &= (x+(3x-2x^3)(1-2x^2+2x^4))(1-x(3x-2x^3)(1-2x^2+2x^4)). \end{align*}Now compute \begin{align*} & 4x(1-x^2)(1-4x^2(1-x^2)^2)-(x+(3x-2x^3)(1-2x^2+2x^4))(1-x(3x-2x^3)(1-2x^2+2x^4)) \\ =& 2x^5(x^2-1)(2x^2-1)(2x^2-3)(2x^4-4x^2+3). \end{align*}Therefore the range of $f$ is contained in $\{0, \pm 1, \pm\sqrt{\tfrac{1}{2}}, \pm\sqrt{\tfrac{3}{2}}\}$. We can check that this makes it impossible for $f(2t)=\pm 1$, so we can remove $\pm 1$ from the range. Next, if $f(x)=\pm\sqrt{\tfrac{1}{2}}$ and $f(y)=\pm\sqrt{\tfrac{3}{2}}$, then $f(x+y)$ is not one of the possible numbers in the range, so the range is contained in either $\{0, \pm \sqrt{\tfrac{1}{2}}\}$ or $\{0, \pm \sqrt{\tfrac{3}{2}}\}$. In the latter case, this results in $f(3t)=0$ always, so $f\equiv 0$. Finally, assume the range of $f$ is contained in $\{0, \pm \sqrt{\tfrac{1}{2}}\}$. We can check that $f(x)=\pm\sqrt{\tfrac{1}{2}}, f(x+y)=\mp \sqrt{\tfrac{1}{2}}$ is impossible in the original FE, so assume that the range is contained in $\{0, \sqrt{\tfrac{1}{2}}\}$ (the other case is similar). Then we have $f(x)= \sqrt{\tfrac{1}{2}}\implies f(x+y)= \sqrt{\tfrac{1}{2}}$ in the original FE, so we only get constant solutions. In summary, there are three solutions: $\boxed{f\equiv 0}, \boxed{f\equiv \sqrt{\tfrac{1}{2}}}, \boxed{f\equiv -\sqrt{\tfrac{1}{2}}}$. (Yes, you can solve this problem without using a degree $15$ polynomial. No, I will not do it.) Thanks to Wolfram Alpha!

You don't need a 15 degree polynomial! Here's how I solved it: The only solutions are $$f \equiv 0, f \equiv \frac{1}{\sqrt{2}} \text{ and } f \equiv -\frac{1}{\sqrt{2}}$$all of which are easily seen to work. Now we show these are the only solutions. Let $P(x,y)$ denote the given assertion. Then $$P(0,0) \Rightarrow 2f(0)(1-f(0)^2)=f(0) \Rightarrow f(0)=0 \text{ or } \pm \frac{1}{\sqrt{2}}$$First consider $f(0) \neq 0$. Note that if $f$ works then $-f$ also works, so we can WLOG assume $f(0)=\frac{1}{\sqrt{2}}$. Then $$P(x,0) \Rightarrow \left(f(x)+\frac{1}{\sqrt{2}} \right) \left(1-\frac{f(x)}{\sqrt{2}} \right)=f(x) \Rightarrow f(x)=\frac{1}{\sqrt{2}} \text{ or } f(x)=\sqrt{2}$$If $f(x)=\sqrt{2}$, then $P(x,x)$ gives $f(2x)=-2 \sqrt{2}$, which is not in the range of $f$. Thus, $f(x)=\frac{1}{\sqrt{2}}$ for all $x \in \mathbb{R}$, as desired. Now suppose $f(0)=0$. Then $$P(x,-x) \Rightarrow f(-x)=-f(x) \text{ or } f(-x)=\frac{1}{f(x)}$$Also, from $P(x,x)$ we have $f(2x)=2f(x)(1-f(x)^2)$. The above two relations can be used as follows- $$P(2x,-x) \Rightarrow (f(2x)+f(-x))(1-f(2x)f(-x))=f(x)$$If $f(-x)=-f(x)$, then taking $z=f(x)$ for brevity, we get $$(2z(1-z^2)-z)(1+2z^2(1-z^2))=z \Rightarrow z(4z^6-6z^4)=0 \Rightarrow f(x)=0 \text{ or } \pm \sqrt{\frac{3}{2}}$$And, if $f(-x)=\frac{1}{f(x)}$, then we have $$\left(2z(1-z^2)+\frac{1}{z} \right)(1-2(1-z^2))=z \Rightarrow (1+2z^2-2z^4)(1-2z^2)=-z^2$$$$\Rightarrow 4z^6-6z^4+z^2+1=0 \Rightarrow (z^2-1)(4z^4-2z^2-1)=0 \Rightarrow f(x)=\pm 1 \text{ or } \pm \frac{\sqrt{\sqrt{5}+1}}{2}$$If $f(x)=\pm \frac{\sqrt{\sqrt{5}+1}}{2}$, then $f(-x)=\frac{1}{f(x)}$ is not in the range of $f$ (the range consists of the $7$ values given above). So $\pm \frac{\sqrt{\sqrt{5}+1}}{2}$ is not in range of $f$. Also, since $f(x)=2f \left(\frac{x}{2} \right) \left(1-f \left(\frac{x}{2} \right)^2 \right),$ and $f \left(\frac{x}{2} \right)$ takes values in $0, \pm 1, \pm \sqrt{\frac{3}{2}}$, so $f(x)$ can never attain the value $\pm 1$ as well. Thus, the image of $f$ is $0, \pm \sqrt{\frac{3}{2}}$ only, and $f$ is an odd function. Finally, if $f(x)=0$, for some $x \in \mathbb{R}$, then $f(2x)=0$, and so $P(2x,x)$ gives $f(3x)=0$. And if $f(x) \neq 0$, then one can easily check that $P(x,x)$ gives $f(2x)=-f(x)$ for the given range of values. Then $P(2x,x)$ again gives $f(3x)=0$. Thus, $f(3x)=0$ for all $x \in \mathbb{R}$, giving the solution $f \equiv 0$. $\blacksquare$

@3 above (Physicsknight), you can't assume continuity and derivability! That makes your solution to $f(0)=0$ case a complete bogus one. @below your solution to case 1 doesn't work, as you can't just compare the coefficients of two polynomials, unless they are equal for more inputs than the degree of the polynomial. It might be possible that the range of $f$ has at most $4$ elements, which is why you need to deal with the solution as above. (This comment was for a solution which has now been deleted)

math_pi_rate wrote: You don't need a 15 degree polynomial! Here's how I solved it: The only solutions are $$f \equiv 0, f \equiv \frac{1}{\sqrt{2}} \text{ and } f \equiv -\frac{1}{\sqrt{2}}$$all of which are easily seen to work. Now we show these are the only solutions. Let $P(x,y)$ denote the given assertion. Then $$P(0,0) \Rightarrow 2f(0)(1-f(0)^2)=f(0) \Rightarrow f(0)=0 \text{ or } \pm \frac{1}{\sqrt{2}}$$First consider $f(0) \neq 0$. Note that if $f$ works then $-f$ also works, so we can WLOG assume $f(0)=\frac{1}{\sqrt{2}}$. Then $$P(x,0) \Rightarrow \left(f(x)+\frac{1}{\sqrt{2}} \right) \left(1-\frac{f(x)}{\sqrt{2}} \right)=f(x) \Rightarrow f(x)=\frac{1}{\sqrt{2}} \text{ or } f(x)=\sqrt{2}$$If $f(x)=\sqrt{2}$, then $P(x,x)$ gives $f(2x)=-2 \sqrt{2}$, which is not in the range of $f$. Thus, $f(x)=\frac{1}{\sqrt{2}}$ for all $x \in \mathbb{R}$, as desired. Now suppose $f(0)=0$. Then $$P(x,-x) \Rightarrow f(-x)=-f(x) \text{ or } f(-x)=\frac{1}{f(x)}$$Also, from $P(x,x)$ we have $f(2x)=2f(x)(1-f(x)^2)$. The above two relations can be used as follows- $$P(2x,-x) \Rightarrow (f(2x)+f(-x))(1-f(2x)f(-x))=f(x)$$If $f(-x)=-f(x)$, then taking $z=f(x)$ for brevity, we get $$(2z(1-z^2)-z)(1+2z^2(1-z^2))=z \Rightarrow z(4z^6-6z^4)=0 \Rightarrow f(x)=0 \text{ or } \pm \sqrt{\frac{3}{2}}$$And, if $f(-x)=\frac{1}{f(x)}$, then we have $$\left(2z(1-z^2)+\frac{1}{z} \right)(1-2(1-z^2))=z \Rightarrow (1+2z^2-2z^4)(1-2z^2)=-z^2$$$$\Rightarrow 4z^6-6z^4+z^2+1=0 \Rightarrow (z^2-1)(4z^4-2z^2-1)=0 \Rightarrow f(x)=\pm 1 \text{ or } \pm \frac{\sqrt{\sqrt{5}+1}}{2}$$If $f(x)=\pm \frac{\sqrt{\sqrt{5}+1}}{2}$, then $f(-x)=\frac{1}{f(x)}$ is not in the range of $f$ (the range consists of the $7$ values given above). So $\pm \frac{\sqrt{\sqrt{5}+1}}{2}$ is not in range of $f$. Also, since $f(x)=2f \left(\frac{x}{2} \right) \left(1-f \left(\frac{x}{2} \right)^2 \right),$ and $f \left(\frac{x}{2} \right)$ takes values in $0, \pm 1, \pm \sqrt{\frac{3}{2}}$, so $f(x)$ can never attain the value $\pm 1$ as well. Thus, the image of $f$ is $0, \pm \sqrt{\frac{3}{2}}$ only, and $f$ is an odd function. Finally, if $f(x)=0$, for some $x \in \mathbb{R}$, then $f(2x)=0$, and so $P(2x,x)$ gives $f(3x)=0$. And if $f(x) \neq 0$, then one can easily check that $P(x,x)$ gives $f(2x)=-f(x)$ for the given range of values. Then $P(2x,x)$ again gives $f(3x)=0$. Thus, $f(3x)=0$ for all $x \in \mathbb{R}$, giving the solution $f \equiv 0$. $\blacksquare$

@3 above (Physicsknight), you can't assume continuity and derivability! That makes your solution to $f(0)=0$ case a complete bogus one. @below your solution to case 1 doesn't work, as you can't just compare the coefficients of two polynomials, unless they are equal for more inputs than the degree of the polynomial. It might be possible that the range of $f$ has at most $4$ elements, which is why you need to deal with the solution as above. (This comment was for a solution which has now been deleted) I have pretty similar solution to you. As for the problem I agree maybe it is a bit easy for P3 but we didn't have so many choices. Also this at IMO could have consider as medium because you will have a lot work to do and calculation espacially if you do the way as a1267ab did have to do a polynomial of degree 15 which actually is not so bad to factorise if you know what you are doing.

Sketch of my solution: Step $1$: Show that $f(0)$ is one of $0, \pm \frac{1}{\sqrt{2}}$. Then use $P(x,0)$ and $P(x,x)$ to conclude that $f(0) \neq 0$ implies $f(x)$ is constant. Step $2$: Consider the case $f(0)=0$. Use $P(x,-x)$ to show that either $f(x)+f(-x)=0$ or $f(x)f(-x)=1$. Step $3$: Use $P(x,x)$ and $P(2x,-x)$ alongwith the result of step $2$ to show that for any real $x$, $f(x)$ is either $0, \pm \sqrt{2}, \pm \frac{1}{\sqrt{2}}$ or a root of $2t^4-3t^2-1=0$. Step $4$: Eliminate the possible values of $f(x)$ using $P(x,x)$ and $P(x,-x)$ again to show that $f(x)$ is one of $0, \pm \frac{1}{\sqrt{2}}$. Step $5$: Thus now check that $f(x)+f(-x)=0$ for all $x$. If $f(x) \neq 0$ for some $x$, use $P(x,0)$ and $P(x,x)$ and $P(2x,-x)$ to derive a contradiction that $f(x)$ has two different values. Thus for all $x$, $f(x)$ must be $0$ if $f(0)=0$. So the only solutions are constant ones $f(x)=0, f(x)=\frac{1}{\sqrt{2}}$ and $f(x)=\frac{-1}{\sqrt{2}}$.

Where's Mr Nemo?( you know who I'm talking about) I thought there were supposed to be official solutions? Or am I wrong or what?