parmenides51 wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying the following condition: for any real numbers $x$ and $y$ , the number $f(x + f(y))$ is equal to $x + f(y)$ or $f(f(x)) + y$

Let $P(x,y)$ be the assertion $f(x+f(y))\in\{x+f(y),f(f(x))+y\}$

Let $U=\{x$ such that $f(x)\ne x\}$
Let $V=\mathbb R\setminus U=\{x$ such that $f(x)=x\}$

If $U=\emptyset$ , we get solution $f(x)=x\quad\forall x$
If $U\ne\emptyset$ , let $u\in U$ :

$P(u-f(x),x)$ $\implies$ new assertion $Q(u,x)$ : $f(u)=f(f(u-f(x)))+x$ $\forall u\in U,\forall x$
This implies $f(x)$ bijective and so $Q(u,f(u))$ , which is $f(f(u-f(f(u))))=0$ , implies $f(f(u))=u+a$ for some constant $a\ne 0$

And so $\forall x$ : either $f(f(x))=x$ (if $x\in V$ ), either $f(f(x))=x+a$ (if $x\in U$
And $Q(u,x)$ implies $f(u)\in\{u-f(x)+x,u-f(x)+x+a\}$ , $\forall x$

If $\exists v\in V$ (so that $f(v)=v$ ):
Setting $x=v$ in previous line, we get $f(u)\in\{u,u+a\}$ and so $f(u)=u+a$
But then $f(f(u))=u+a$ is $f(u+a)=u+a=f(u)$ , impossible since injective
And so $V=\emptyset$ and $f(f(x))=x+a$ $\forall x$

Then $Q(u,x)$ is $f(u)=u-f(x)+a+x$ $\forall x$ and so $f(x)=x+c$ , which indeed fits, whatever is $c\ne 0$

And solution (grouping the two cases) : $\boxed{f(x)=x+c\quad\forall x}$ , which indeed fits, whatever is $c\in\mathbb R$