1. The function $f$ is defined for every real number and for any $x$ and $y$ the following holds:

$$
f(x y)=f(x) \cdot f(y)-f(x+y)+1
$$

where $f(1)=2$. Determine $f(m)$ for every integer $m$.
1. From $f(1 \cdot m)=f(1) f(m)-f(1+m)+1$ it follows that

$$
f(m)=2 f(m)-f(m+1)+1 \quad \text { i.e. } \quad f(m+1)=f(m)+1
$$

Now we get $f(2)=f(1)+1=3, f(3)=f(2)+1=4$, $f(4)=f(3)+1=5, \ldots$ We assume that $f(m)=m+1$ for every natural number $m$. This can be proven by mathematical induction.

$1^{\circ} \mathrm{For} m=1$ the statement holds, since $f(1)=2=1+1$.

$2^{\circ}$ Assume that for some $m \geq 1$ it holds that $f(m)=m+1$.

$3^{\circ}$ Prove that $f(m+1)=m+2$. From $f(m)=m+1$ it follows that

$$
f(m+1)=f(m)+1=m+1+1
$$

Thus, the statement is proven for every natural number $m$.

15 points

From $(*)$ for $m=0$ it follows that $f(0)=f(1)-1=1$.

From $(*)$ for $m=-1$, we get $f(0)=f(-1)+1$, i.e.

$f(-1)=f(0)-1=0$.

Now let $m$ be a natural number. Then:

$$
\begin{aligned}
f(m-1) & =f((-1) \cdot(-m+1)) \\
& =f(-1) f(-m+1)-f(-m)+1 \\
& =-f(-m)+1
\end{aligned}
$$

from which, due to $f(m-1)=m$, we get

$$
\therefore f(-m)=-f(m-1)+1=-m+1
$$

which means that the statement holds in this case as well.

10 points

Therefore, for every integer $m$ it is $f(m)=m+1$.