qweDota wrote: Find all real numbers $a$ such that there exist $f:\mathbb{R} \to \mathbb{R}$ with $$f(x-f(y))=f(x)+a[y]$$for all $x,y\in \mathbb{R}$ Let $P(x,y)$ be the assertion $f(x-f(y))=f(x)+a\lfloor y\rfloor$ Obviously $a=0$ has such "$f$" (choose for example constant $f$). So let us consider from now that $a\ne 0$ If $f(u)=f(v)$ for some $u,v$, comparing $P(x,u)$ with $P(x,v)$ immediately gives $\lfloor u\rfloor=\lfloor v\rfloor$ If $f(u)\notin\mathbb Z$, then $\exists x_1,x_2$ such that $\lfloor x_1\rfloor=\lfloor x_2\rfloor$ but $\lfloor x_1-f(u)\rfloor\ne\lfloor x_2-f(u)\rfloor$ But $P(x_1,u)$ and $P(x_2,u)$ imply then $f(x_1-f(u))=f(x_2-f(u))$ which implies $\lfloor x_1-f(u)\rfloor=\lfloor x_2-f(u)\rfloor$ Contradiction So $f(x)\in\mathbb Z$ $\forall x$ Simple induction using $P(x,y)$ gives $f(x-nf(y))=f(x)-an\lfloor y\rfloor$ $\forall n\in\mathbb Z$ So $f(x-f(z)f(y))=f(x)-af(z)\lfloor y\rfloor$ $\forall x,y,z$ Swapping $y,z$ and subtracting, we get $f(z)\lfloor y\rfloor=f(y)\lfloor z\rfloor$ $\forall y,z$ Setting there $z=1$, we get $f(x)=c\lfloor x\rfloor$ $\forall x$ and for some $c\in\mathbb Z$ Plugging this back in original equation, we get $a=-c^2$ Hence the answer $\boxed{a\in\{-n^2\quad\forall n\in\mathbb Z\}}$

qweDota wrote: Find all real numbers $a$ such that there exist $f:\mathbb{R} \to \mathbb{R}$ with $$f(x-f(y))=f(x)+a[y]$$for all $x,y\in \mathbb{R}$ See also here https://artofproblemsolving.com/community/c6h1574311p9683121

pco wrote: If $f(u)\notin\mathbb Z$, then $\exists x_1,x_2$ such that $\lfloor x_1\rfloor=\lfloor x_2\rfloor$ but $\lfloor x_1-f(u)\rfloor\ne\lfloor x_2-f(u)\rfloor$ But $P(x_1,u)$ and $P(x_2,u)$ imply then $f(x_1-f(u))=f(x_2-f(u))$ which implies $\lfloor x_1-f(u)\rfloor=\lfloor x_2-f(u)\rfloor$ Contradiction If $\lfloor x_1\rfloor=\lfloor x_2\rfloor$ that dont mean that $f(x_1)=f(x_2)$

helol wrote: If $\lfloor x_1\rfloor=\lfloor x_2\rfloor$ that dont mean that $f(x_1)=f(x_2)$ It does. If $\lfloor u\rfloor=\lfloor v\rfloor$, we have $f(x-f(u))=f(x-f(v))$ and so $\lfloor x-f(u)\rfloor=\lfloor x-f(v)\rfloor$ $\forall x$ and so $f(u)=f(v)$