Korean Mathematical Olympiad Final Round

Put $x=0$ in the given equation $$f(x^{2015}+f(y{)}^{2015})=f(x{)}^{2015}+y^{2015}.(1)$$ Then, we have $$f(f(y){)}^{2015}=f(0{)}^{2015}+y^{2015}.(2)$$ It implies that $f$ is a bijective function. By putting $f(x)$ to $x$ in (1), we have $$f(f(x){)}^{2015}+f(y{)}^{2015}=f(f(x){)}^{2015}+y^{2015}(3)$$ and by putting $f(y)$ to $y$ in (1), $$f(x^{2015}+f(f(y){)}^{2015})=f(x{)}^{2015}+f(y{)}^{2015}.(4)$$ In (3), we swipe $x$ and $y$ first and then take $f$ on both sides, then we have $$f(f(f(x){)}^{2015}+f(y{)}^{2015}))=f(x^{2015}+f(f(y){)}^{2015})=f(x{)}^{2015}+f(y{)}^{2015}.$$ Since $f$ is bijective, we have $$f(f(x))=x.\text{\hspace{1em}(5)}$$ By (2), (3), and (5), we have $$(f(f(x){)}^{2015}+f(y{)}^{2015})=x^{2015}+y^{2015}=f(f(x{)}^{2015})+f(f(y{)}^{2015})-2f(0{)}^{2015}$$ and, since $f$ is bijective, $f(x+y)=f(x)+f(y)-2f(0{)}^{2015}$. Here, by putting $x=y=0$, we have $f(0)=2f(0{)}^{2015}$. Hence $f(0)$ is $0$ or $\pm (\frac{1}{2}{)}^{\frac{1}{2014}}$. By putting $f(y)$ to $y$ in (2), we get $f(y^{2015})=f(0{)}^{2015}+f(y{)}^{2015}$. Consider an equation $x^{2015}-x+f(0{)}^{2015}$. It must have three different real roots $f(-1)$, $f(0)$ and $f(1)$. However, if $f(0)=\pm (\frac{1}{2}{)}^{\frac{1}{2014}}$, this equation does not have three different real roots. Therefore, $f(0)=0$ and $f(1)$ is either $1$ or $-1$. Now we have $f(x+y)=f(x)+f(y)$ and $f(y^{2015})=f(y{)}^{2015}$, and $f(1)$ is either $1$ or $-1$. We will prove that only $f(x)=x$ and $f(x)=-x$ satisfy these three conditions. If $f(1)=-1$, then we may consider $g(x)=-f(x)$. It allows us to consider only the case where $f(1)=1$. By the general Cauchy method, we have $f(x)=x$ for all $x\in \mathbb{Q}$. Now we need to extend this function to real numbers. To do that, we will prove that $f(x^2)=f(x{)}^2$. Let $\left(\genfrac{}{}{0ex}{}{2015}{k}\right)f(x^k)-f(x{)}^k=x_{2015-k}$ for $k=0,1,2,\dots ,2015$. Then, for any integer $m$, by calculating $f((x+m{)}^{2015})=(f(x)+f(m){)}^{2015}$, we have $$\sum _{i=0}^{2015}{m}^i{x}_i=0.$$ Now we think of it as a system of infinite number of equations. Claim. The system has a unique solution $(x_0,x_1,\dots ,x_{2015})=(0,0,\dots ,0)$. Proof. Suppose there is another solution $(y_0,y_1,\dots ,y_{2015})=(0,0,\dots ,0)$. Let $s$ be the maximum index with $y_s\ne 0$. By multiplying $\frac{1}{y_s}$, we may assume that $y_s=1$. Set $m$ as a number bigger than $|y_0|+|y_1|+\cdots +|y_{2015}|$. Then, $$0=\sum _{i=0}^{2015}{m}^i{y}_i=\sum _{i=0}^s{m}^i{y}_i>m^s{y}_s-\sum _{i=0}^{s-1}{m}^i|y_i|\ge m^s-m^{s-1}\sum |y_i|>0$$ which yields a contradiction. Therefore it has a unique solution $(0,0,\dots ,0)$. $\square$

The above claim says that $f(x^k)=f(x{)}^k$ for $k=0,1,2,\dots ,2015$. Especially, $f(x^2)=f(x{)}^2\ge 0$. Now, for every $x>y$, we have $f(x)-f(y)=f(x-y)=f({\sqrt{x-y}}^2)\ge 0$, so $f$ is increasing. Therefore, $f(x)=x$, and the candidates of the original problem are $f(x)=x$ and $f(x)=-x$. Indeed, one can check that they are real solutions of the problem by substituting.