20.1. [10.7 (15 points)] The function $f$, defined on the set of integers, satisfies the following conditions:

1) $f(1)+1>0$
2) $f(x+y)-x f(y)-y f(x)=f(x) f(y)-x-y+x y$ for any $x, y \in \mathbb{Z}$;
3) $2 f(x)=f(x+1)-x+1$ for any $x \in \mathbb{Z}$.

Find $f(10)$.
Answer: 1014.

Solution. If $h(x)=f(x)+x$, then from condition 2) we get $h(x+y)=h(x) h(y)$. Then, for $x=y=0$, this equality takes the form $h(0)^{2}=h(0)$, i.e., $h(0)=0$ or $h(0)=1$. In the first case, $h(x) \equiv 0$, which is impossible due to condition 1). If $a=h(1)$, then $h(x)=a^{x}$ for any $x \in \mathbb{Z}$. From condition 3), we find $a=2$, then $h(10)=2^{10}=1024, f(10)=h(10)-10=1014$.