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 $x,y$ .

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.

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