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 $c\in\mathbb R$

The only function that satisfy are $\boxed{f(x)=x+a}$ for any real number $a$, which can be directly verified to work. We now show that these are the only solutions. Let $P(x,y)$ denote the assertion. Let $f(0)=a$. Claim: $f$ is bijective If $y_1\neq y_2$ and $f(y_1)=f(y_2)$ then comparing the assertions $P(x,y_1)$ with $P(x,y_2)$ gives that we must have $f(x+f(y_1))=x+f(y_1)$ which must mean $f$ is just the identity. Thus $f$ is injective. Then the assertion $P(x,0)$ gives that $f(x+a)=x+a$ or $f(x+a)=f(f(x))$ but as $f$ is injective the last one is equivalent to $f(x)=x+a$, thus $x+a$ is in the domain of $f$ for all $x$, making $f$ surjective. The assertion $P(f(z)-a,0)$ gives that either $f(f(z))=f(z)$ or $f(f(z))=f(f(f(z)-a))$ but as $f$ is injective we must have either $f(z)=z$ or $f(z)=z+a$. If $a=0$ then we are done so assume $a\neq 0$. If $f(x)=x$ and $f(w)=w+a$ then chose $y$ such that $f(y)=w-x$, yielding a contradiction for $P(x,y)$. This finishes the proof.