1999 IMO Problems/Problem 6

Problem

Determine all functions $f:\Bbb{R}\to \Bbb{R}$ such that

\[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]

for all real numbers $x,y$.

Solution

Let $f(0) = c$. Substituting $x = y = 0$, we get:

\[f(-c) = f(c) + c - 1. \hspace{1cm}    ... (1)\] Now if c = 0, then:

\[f(0) = f(0) - 1\] which is not possible.

$\implies c \neq 0$.

Now substituting $x = f(y)$, we get

\[c = f(x) + x^{2} + f(x) - 1\].

Solving for $f(x)$, we get \[f(x) = \frac{c + 1}{2} - \frac{x^{2}}{2}. \hspace{1cm}  ... (2)\]

This means $f(x) = f(-x)$ because $x^{2} = (-x)^{2}$.

Specifically, \[f(c) = f(-c). \hspace{1cm}    ... (3)\]

Using equations $(1)$ and $(3)$, we get:

\[f(c) = f(c) + c - 1\]

which gives

\[c = 1\].

So, using this in equation $(2)$, we get

\[\boxed{f(x) = 1 - \frac{x^{2}}{2}}\] as the only solution to this functional equation.

See Also

1999 IMO (Problems) • Resources
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