Find function $f:\Bbb{R} \to \Bbb{R}$ such that
$$\{f(x)\}^2 +2yf(x)+f(y)=f(y+f(x))$$where $\{ f(x) \}$ is the fractional part of $f(x)$
This is an upgrade from JMO Finals 2006. The original problem was $f(x)^2+2y\cdot f(x)+f(y)=f(y+f(x))$ for $x,y \in \Bbb{R}$. I solved the original problem by using the replacement technique:
Suppose that $f(x)\neq0$ for $x \in \Bbb{R}$. So there exists a real so that $f(a)=b\neq0$. Replacing $x$ with $a$, we get
$f(y+b)-f(y)=b^2+2yb\tag{1}\label{1}$
Next, replace $x$ with $y+b$, $y$ with $-f(y)$, we get
$f(f(y+b)-f(y))= (f(y+b)-f(y))^2+f(-f(y))-f(y)^2 \tag{2} \label{2}$
Notice that $f(-f(y))=f(y)^2+f(0)\tag{3}\label{3}$ so from $(\ref{1}), (\ref{2}), (\ref{3})$, we get
$f(b^2+2yb)=(b^2+2yb)^2+f(0)$
$\Rightarrow f(x)=x^2+c$ with $x \in \Bbb{R}$ and constant $c$.
This is my solution for the original problem. I have no clue for the upgrade one so I’m grateful for any ideas.