$(x-y)(f(f(x)^2)-f(f(y)^2))=(f(x)-f(y))(f(x)^2-f(y)^2$

Question

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}

Could you tell the origin of this problem ? A 2015 Olympiad somewhere on the globe ? — Jean Marie, Jan 19 '22 at 08:05
@RamanujanXV My expectation is that $f(x) = 2015x$ or $f(x)=2015 \sqrt{x}.$ — RDK, Jan 20 '22 at 01:49
@JeanMarie I just found it on the problem collection, which has collected various kinds of Olympiad Problems. — RDK, Jan 20 '22 at 01:51
Same questions (with $2013$ instead of $2015$) on AOPS: "(x-y)(f(f^2(x))-f(f^2(y)))=(f(x)-f(y))(f^2(x)-f^2(y))" and "function equation". I found them using Approach Zero — Mohsen Shahriari, Jan 24 '22 at 01:09
Please avoid titles that only consist of Mathjax, and edit the current title by adding some text (a parenthesis is also missing). Also, please search before posting, to avoid duplication. Approach Zero works very well for functional equation post on MSE and AOPS. Those aside, one of the AOPS posts linked above suggests that the source is Problem 5 from day 2 of Vietnam Mathematical Olympiad 2013. You can edit your post and add this (after verifying it, if you really want it to be correct). — Mohsen Shahriari, Jan 24 '22 at 01:14

score 0 · Accepted Answer · answered Apr 10 '22 at 02:35

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}

$(x-y)(f(f(x)^2)-f(f(y)^2))=(f(x)-f(y))(f(x)^2-f(y)^2$

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