The 16th Japanese Mathematical Olympiad - The Final Round

The given equation clearly holds if $f(x)=0$ for all $x$. Assume that there exists $a$ such that $f(a)\ne 0$. Substituting $y=-f(x)$ into the given equation and letting $c=f(0)$, we obtain $$f(-f(x))=c+f(x{)}^2.(1)$$ Substituting $y=-f(y)$ into the equation and using (1), $$\begin{array}{rl}f(f(x)-f(y))& =f(x{)}^2-2f(x)f(y)+f(-f(y))\\ & =f(x{)}^2-2f(x)f(y)+f(y{)}^2+c\\ & =(f(x)-f(y){)}^2+c.\end{array}(2)$$

$$f(a{)}^2+2yf(a)=f(y+f(a))-f(y).$$ Since the left-hand side runs through $\mathbb{R}$ when $y$ runs through $\mathbb{R}$, the right-hand side runs through $\mathbb{R}$. Hence, $f(x)-f(y)$ runs all real numbers when $x$ and $y$ run through $\mathbb{R}$. Therefore, according to (2), $f(x)=x^2+c$ for all $x$. This meets the given equation for any constant $c$. Therefore, $f(x)=x^2+c$ ($c$ any constant) or $f(x)=0$ (for all $x$).