Selection and Training Session

Answer: there are no such functions.

Suppose that there exist functions $f,g$ satisfying the equality $$f(x+f(y))=y^2+g(x).(\ast )$$ 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 $$z^2=y^2.(1)$$ 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 $$f(-y)=f(y).(2)$$ Set $f(0)=a$. Putting $y=0$ in (), we have $$g(x)=f(x+a).(3)$$ Now, $$\begin{array}{rl}f(x+f(y))& =y^2+f(x+a)=y^2+f(-x-a)=\\ & =y^2+f((-x-2a)+a)=f(-x-2a+f(y)).\end{array}$$ Hence from (1) it follows that $$(x+f(y){)}^2=(-x-2a+f(y){)}^2\iff (2f(y)-2a)(2x+2a)=0$$ 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 $g$.