parmenides51 wrote: Do there exist functions $f$ and $g$, $f : R \to R$, $g : R \to R$ such that $f(x + f(y)) = y^2 + g(x)$ for all real $x$ and $y$ Let $a=f(0)$ $f(x+f(y))=y^2+g(x)$ Setting there $y=0$, we get $g(x)=f(x+a)$ And so assertion $P(x,y)$ : $f(x+f(y))=y^2+f(x+a)$ Comparing $P(x,y)$ with $P(x,-y)$, we get $f(x+f(y))=f(x+f(-y))$ And so $f(x)=f(x+f(y)-f(-y))$ $\forall x,y$ Comparing then $P(x,y)$ with $P(x,y+f(z)-f(-z))$, we get $(y+f(z)-f(-z))^2=y^2$ $\forall y,z$ And so $f(z)=f(-z)$ and $f(x)$ is even. $P(-f(x),x)$ $\implies$ $f(a-f(x))=a-x^2$ $P(-a,x)$ $\implies$ $f(f(x)-a)=a+x^2$ And so, since even, $a-x^2=a+x^2$ $\forall x$, impossible. And so $\boxed{\text{No such function}}$