Belarus — Selection and Training Session

Answer: there are no such functions.

Suppose that there exist functions $f, g$ satisfying the equality
$$f(x + f(y)) = y^2 + g(x). \quad (*)$$
First, suppose that $f(y) = f(z)$ for some $y, z \in \mathbb{R}$. Then $z^2 + g(x) = f(x + f(z)) = f(x + f(y)) = y^2 + g(x)$, whence
$$z^2 = y^2. \quad (1)$$
Now, $f(x+f(y)) = y^2 + g(x) = (-y)^2 + g(x) = f(x+f(-y))$. Using (1), we get $(x+f(y))^2 = (x+f(-y))^2$ for all $x, y$. That is $x+f(y) = -x-f(-y)$, or $x+f(y) = x+f(-y)$. Of these two equalities, the former is impossible since the equality $2x = -f(-y) - f(y)$ implies that $2x$ is a constant which is not true. Hence
$$f(-y) = f(y). \qquad (2)$$
Set $f(0) = a$. Putting $y = 0$ in (*), we have
$$g(x) = f(x + a). \qquad (3)$$
Now,
$$\begin{aligned} f(x + f(y)) &= y^2 + f(x + a) = y^2 + f(-x - a) = \\ &= y^2 + f((-x - 2a) + a) = f(-x - 2a + f(y)). \end{aligned}$$
Hence from (1) it follows that
$$(x + f(y))^2 = (-x - 2a + f(y))^2 \Leftrightarrow (2f(y) - 2a)(2x + 2a) = 0$$
for all $x, y \in \mathbb{R}$, which implies $f(y) = a$, and (3) gives $g(x) = a$. So, (*) becomes $a = y^2 + a$ for all $y \in \mathbb{R}$, a contradiction. Therefore there are no such functions $f$ and $g$.