Find $f: \mathbb{R} \to \mathbb{R}$ which satisfies:
$f\small(0)\normalsize=0, f\small(1)\normalsize=2015. \\
(x-y)(f\small(f(x)^2\small)\normalsize-f\small(f(y)^2\small)\normalsize)=(f\small(x)\normalsize-f\small(y)\normalsize)(f\small(x)^2\normalsize-f\small(y)^2\normalsize)$
My Attempt:
\begin{align}
&f \not\equiv c. \\
\Rightarrow \; & \color{blue} {\exists \ t_k \text{ s.t. } f(t_k) \neq k.} & \tag{1} \label{1} \\
& \color{red} {\exists y_1, y_2 \text{ s.t. } f(y_1) \neq f(y_2).} \tag{2} \label{2} \\
\ \\
&\text{let } f(a)=f(b), f(x) \neq f(a) \text{ by } \color{red}{(\ref 2)}. \\
P(x, a): \; & (x-a)(f\small(f(x)^2)\normalsize-f(f\small(a)^2\normalsize))=(f(x)-f(a))(f(x)^2-f(a)^2). \\
P(x, b): \; & (x-b)(f\small(f(x)^2)\normalsize-f(f\small(b)^2\normalsize))=(f(x)-f(b))(f(x)^2-f(b)^2). \\
& f(x) \neq f(a), f(x) \neq f(b). \\
\therefore \; & (f(x)-f(a))(f(x)^2-f(a)^2) = (f(x)-f(b))(f(x)^2-f(b)^2) \neq 0. \\
\Rightarrow \; & (x-a)(f\small(f(x)^2)\normalsize-f\small(f(a)^2)\normalsize)=(x-b)(f\small(f(x)^2)\normalsize – f\small(f(b)^2)\normalsize). \\
& f\small(f(x)^2)\normalsize – f\small(f(a)^2)\normalsize) = f\small(f(x)^2)\normalsize – f\small(f(b)^2)\normalsize) \ \Rightarrow \ x-a=x-b, a=b. \\
\therefore \; & \color{green}{f(a)=f(b) \Rightarrow a=b.} \label{3} \tag{3}
\end{align}
Best Answer
Posted on AOPS, and I am posting the arranged solution.
\begin{align} P(x, y): \; & (x-y)(f(f(x)^2)-f(f(y)^2))=(f(x)-f(y))(f(x)^2-f(y)^2).\\ P(x, 0): \; & xf(f(x)^2)=f(x)^3. \\ \ \\ \therefore \; & yf(f(x)^2)+xf(f(y)^2)=f(x)f(y)(f(x)+f(y)). \text{ (Let this one be $Q(x, y).$)} \\ \ \\ Q(1, 1): \; & f(2015^2)=2015^3. \\ \ \\ Q(1, x): \; & 2015^3x+f(f(x)^2)=2015f(x)(2015+f(x)). \\ Q(x, x): \; & xf(f(x)^2)=f(x)^3. \\ xQ(1, x)-Q(x, x): \; & 2015^3x^2=2015^2xf(x)+2015xf(x)^2-f(x)^3. \\ \ \\ \therefore \; & f(x)^3-(2015x)f(x)^2-(2015^2x)f(x)+2015^3x^2=0. \\ \Rightarrow \; & (f(x)-2015x)(f(x)^2-2015^2x)=0. \\ \ \\ \therefore \; & f(x)=2015x \text{ for } \forall x \leq 0. \ (\because f(x)^2-2015^2x >0.) \\ \ \\ P(x, 0): \; & xf(f(x)^2)=f(x)^3. \\ x<0: \; & xf(2015^2x^2)=2015^3x^3, f(2015^2x^2)=2015^3x^2. \\ \Rightarrow \; & f(x)=2015x \text{ for } \forall x \geq 0. \\ \ \\ \therefore \; & f(x)=2015x. \end{align}