INAMO 2023/2 wrote: Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor \]for any $x,y \in \mathbb{R}$. Mine! Thought about this problem when I was thinking back about ISL 2022 A6 and proposed this as a cute trivial problem, but somehow got into the contest Here's the official solution that I submitted. We claim that all such solutions for the above functional equation are of the form $\boxed{f(x) = \lfloor x \rfloor + k}$ for all $x$, for some fixed $k \in \mathbb{Z}$. To see this, we see that for all $x,y \in \mathbb{R}$, \[ f(f(x) + y) = \lfloor \lfloor x \rfloor + y \rfloor + 2k = \lfloor x \rfloor + \lfloor y \rfloor + 2k = \lfloor x + \lfloor y \rfloor + 2k \rfloor = \lfloor x + f(f(y)) \rfloor \]We'll now prove that there are no other solutions. Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation. Claim 01. $\text{Im}(f) \subseteq \mathbb{Z}$. Proof. Fix $\alpha \in \mathbb{R}$. $P(0,\alpha - f(0))$ gives us \[ f(\alpha) = f(f(0) + \alpha - f(0)) = \lfloor f(f(\alpha - f(0))) \rfloor \in \mathbb{Z} \] Claim 02. $f(a) = f(b) \implies \lfloor a \rfloor = \lfloor b \rfloor$. Proof. As $f(f(0)) \in \mathbb{Z}$ from the first claim, $P(a,0)$ and $P(b,0)$ gives us \[ \lfloor a \rfloor + f(f(0)) = f(f(a)) = f(f(b)) = \lfloor b \rfloor + f(f(0)) \implies \lfloor a \rfloor = \lfloor b \rfloor \] To finish this problem, just note that \[ f(f(0) + x) \stackrel{P(0,x)}{=} \lfloor f(f(x)) \rfloor \stackrel{\text{Claim 01}}{=} f(f(x)) \stackrel{\text{Claim 02}}{\implies} f(x) \stackrel{\text{Claim 01}}{=} \lfloor f(x) \rfloor = \lfloor x + f(0) \rfloor \stackrel{\text{Claim 01}}{=} \lfloor x \rfloor + f(0) \]as desired.

A bit more convoluted...

I think everybody I know solved this problem, but anyway here is a short solution. Let $P(x,y)$ be the assertion, clearly $P(x,x-f(x))$ gives us $f(x) \in \mathbb{Z} \quad \forall x \in \mathbb{R}$. $P(x-f(x),0)$ gives us $$f(x-f(x))=\lfloor x \rfloor-f(x)+f(f(0))$$By substituting it to $P(x,x-f(x))$, we will get $$f(x)=2\lfloor x\rfloor -f(x)+f(f(0))$$From here, it's already clear what the answer is.

I was pretty surprised that the FE question reappeared this year, coupled with a bit of seasoning of the floor function, lol. Let $P(x,y)$ be the assertation of \[ f(f(x) + y) = \lfloor x + f(f(y)) \rfloor. \]From $P(x,0)$, we have $f(f(x)) = \lfloor x +c\rfloor$ where $c=f(f(0))$ i.e. \[f(f(x)+y) = \lfloor x + f(f(y))\rfloor = \lfloor x +\lfloor y+c\rfloor \rfloor = \lfloor x \rfloor + \lfloor y+c\rfloor.\]From $P(x,0)$ we must have $\lfloor x+c\rfloor = \lfloor x\rfloor +\lfloor c\rfloor\implies \lfloor \{x\}+\{c\}\rfloor =0$ for all $x\in \mathbb{R}$. If $c\not\in \mathbb{Z}$, take $x=1-\{c\}$ and we have contradiction. Because $c\in \mathbb{Z}$, we have $f(f(x)+y) = \lfloor x \rfloor + \lfloor y\rfloor +c$. From $P(0,-f(0))$ we have $f(0) = \lfloor -f(0)\rfloor + c\implies f(0)\in \mathbb{Z}\implies f(0)=\frac{c}{2}=k\in \mathbb{Z}$. From $P(0,x-f(0))$ we have \[f(x) = \lfloor x-f(0)+c\rfloor = \lfloor x-k+2k\rfloor = \lfloor x \rfloor +k,\;k\in \mathbb{Z}\]and we can check if this solution to the equation.

Another one... Denote $P(x,y)$ as the assertion of $x$ and $y$ to the given functional equation on the problem. From $P(x, x - f(x))$, we obtain $f(x) = \lfloor \text{something} \rfloor$, so $\text{Im}(f) \subseteq \mathbb{Z}$. From $P(f(x), 0)$ we obtain \[ f(f(f(x))) = \lfloor f(x) + f(f(0)) \rfloor = f(x) + f(f(0)) \text{ }(\star) \]And therefore, from comparing $P(x, f(y))$ and $P(y, f(x))$ we obtain: \begin{align*} \lfloor x + f(f(f(y))) \rfloor = f(f(x) + f(y)) = \lfloor y + f(f(f(x))) \rfloor &\stackrel{\text{Im}(f) \subseteq \mathbb{Z}}{\implies} \lfloor x \rfloor + f(f(f(y))) = \lfloor y \rfloor + f(f(f(x))) \\ &\stackrel{(\star)}{\implies} \lfloor x \rfloor + f(y) + f(f(0)) = \lfloor y \rfloor + f(x) + f(f(0)) \\ &\implies f(x) - \lfloor x \rfloor = K, \end{align*}for some constant $K$. Since $f(x)$ and $\lfloor x \rfloor$ are both integers, then $K$ must be an integer as well, so the solutions are of the form $\boxed{f(x)=\lfloor x \rfloor + K}$, where $K \in \mathbb{Z}$. It should be easy to check this into the original equation.

Let $P(x,y)$ be the assertion $P(x,y-f(x)) = f(y) = \lfloor{x+f(f(y-f(x))}\rfloor$ for each real y the function is integer $P(x-f(f(y)),y) = f(f(x-f(f(y)))+y) = \lfloor{x}\rfloor$ this implies f surjective over integer so we can pick u integer so that $f(u) = 0$ proof le $x = 0$ and $y \in \mathbb{Z}$ then we are done $P(x,u) = f(f(x)+u) = u+f(0)$ $P(u,u) = 0 = u+f(0)$ $u = -f(0)$ $P(u,y) = f(y) = u+f(f(y))$ $P(-f(0),y) = f(y)+f(0) = +f(f(y))$ (1) Rewrite the main equation $P(x,y) = f(f(x)+y) = \lfloor{x}\rfloor+f(y)+f(0)$ $P(x,0) = f(f(x)) = \lfloor{x}\rfloor+2f(0) = f(x)+f(0)$ (From equation (one) we get ) $f(x) = \lfloor{x}\rfloor +f(0)$