10.4. Find $f(2021)$ if for any real $x$ and $y$ the equality $f(x+f(y))=x+y$ holds.
Answer: 2021.

Solution. Let $f(0)=b$, then for $y=0$ we get $f(x+b)=x$, from which $f(x)=x-b$. Thus, $f(x+f(y))=f(x+(y-b))=f(x+y-b)=x+y-b-b=x+y-2b$.

Since $f(x+f(y))=x+y$ for any real $x$ and $y$, then $b=0$, so $f(x)=x$, and $f(2021)=2021$.

Comment. Only the example of the function $f(x)=x-1$ gets a point.