parmenides51 wrote: Find all functions $f: R\to R$ such that for all real numbers $x, y$ is satisfied that $$f (x + y) = (f (x))^{ 2013} + f (y).$$ Let $P(x,y)$ be the assertion $f(x+y)=f(x)^{2013}+f(y)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x)^{2013}-f(x)=0$ and so $f(x)\in\{-1,0,1\}$ $\forall x$ If $f(x)=1$ for some $x$, then $P(x,x)$ $\implies$ $f(2x)=2$, impossible If $f(x)=-1$ for some $x$, then $P(x,x)$ $\implies$ $f(2x)=-2$, impossible And so $\boxed{f(x)=0\quad\forall x}$, which indeed fits.

Denote the assertion by $P(x,y).$ By $P(0,0)$ we have $f(0)=0$ and by $P(x,0)$ we have $f(x)=f(x)^{2013}$ so $f(x)\in \{-1,0,1\}.$ And pointwise is easy and we find $f\equiv 0$ only.