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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.

José Carlos Santos
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user1052563
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  • also the "simplified" solve $y=0$ then $f(f(x))=f(0)+f(x)^2$ leading to $f(z)=f(0)+z^2$ doesn't seem to extend well as $f(z)=f(0)+{z}^2$ does not appear to be solution for other $y$. – zwim Oct 22 '22 at 10:23
  • By replacing $y$ with $f(y)$ RHS is symmetric, so LHS must be too, which leads to ${f(x)}^2-f(f(x))=C$ being a constant. That transforms the original equation into $f(f(x))+C+2yf(x)+f(y)=f(y+f(x))$. – Sil Oct 23 '22 at 16:48
  • If $f$ is continuous I think it's easy to prove that the only solution is the zero function. – Jkbb Feb 26 '23 at 19:17

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