3. Find all functions $f: \mathbb{Q} \rightarrow \mathbb{R}$ that satisfy the following conditions:

(1) $f(x+y) - y f(x) - x f(y) = f(x) f(y) - x - y + x y$ for all $x, y \in \mathbb{Q}$;

(2) $f(x) = 2 f(x+1) + 2 + x$ for all $x \in \mathbb{Q}$;

(3) $f(1) + 1 > 0$.

(Kunap)
3. After rearrangement, condition (1) becomes $f(x+y)+x+y=(f(x)+x)(f(y)+y)$. Thus, by substituting $g(x)=f(x)+x$, conditions (1)-(3) become

$$
g(x+y)=g(x) g(y), \quad g(x+1)=\frac{1}{2} g(x), \quad g(1)>0
$$

Further, from $g(1)=g(1) g(0)$ we get $g(0)=1$, and from $g(0)=g(x) g(-x)$ we get $g(x) \neq 0$. Notice that $g(x)=g\left(\frac{x}{2}\right)^{2}>0$ for all $x$. If we set $h(x)=\log _{2} g(x)$, conditions (*) give

$$
\text { (a) } h(x+y)=h(x)+h(y) \quad \text { and } \quad \text { (b) } h(x+1)=h(x)-1 \text {. }
$$

Thus, the function $h$ satisfies Cauchy's equation (a), from which $h(x)=x \cdot h(1)$ for $x \in \mathbb{Q}$, and (b) gives $h(1)=h(0)-1=-1$. Therefore, $h(x)=-x$ and $f(x)=2^{-x}-x$.

It is easy to verify that this function satisfies conditions (1)-(3).