20.6. (MMS, Luxembourg, 80). Find all functions $f: \mathbf{Q} \rightarrow \mathbf{Q}$, satisfying the condition $f(1)=2$ and the identity

$$
f(x y) \equiv f(x) f(y)-f(x+y)+1, \quad x, y \in \mathbf{Q}
$$
20.6. Putting \( y=1 \) in the original identity, we get the identity

\[
f(x) \equiv f(x) f(1)-f(x+1)+1 \quad (x \in \mathbb{Q})
\]

i.e.

\[
f(x+1) \equiv f(x)+1
\]

From this, for all \( x \in \mathbb{Q}, n \in \mathbb{Z} \) we have

\[
f(x+n)=f(x)+n
\]

therefore

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

Next, substituting into the original identity the values \( x=1 / n, y=n \) for \( n \in \mathbb{Z} \), we get

\[
f\left(\frac{1}{n} \cdot n\right)=f\left(\frac{1}{n}\right) \cdot f(n)-f\left(\frac{1}{n}+n\right)+1
\]

from which

\[
2=f(1 / n)(n+1)-f(1 / n)-n+1, \text{ i.e. } f(1 / n)=1+1 / n
\]

Finally, substituting the values \( x=p, y=1 / q \) for \( p \in \mathbb{Z}, q \in \mathbb{N}_{1} \) we have

\[
f\left(p \cdot \frac{1}{q}\right)=f(p) \cdot f\left(\frac{1}{q}\right)-f\left(p+\frac{1}{q}\right)+1
\]

from which

\[
f(p / q)=(p+1)(1 / q+1)-1 / q-p=p / q+1 .
\]

Thus, we have only one function \( f(x)=x+1 \), which indeed satisfies all the conditions of the problem. Then