Find the functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that $f(1)=2, \forall x, y \in \mathbb{Q}: f(x y)=f(x) f(y)-$ $f(x+y)+1$

## - Solution -
By setting $y=1$ we have $f(x+1)=f(x)+1$. An immediate induction shows that $f(x+n)=f(x)+n$ for all integers $n$. Since $P(1)=2, f(m)=m+1$ for all integers $m$. Let $x=\frac{m}{n}$ be a rational number. By setting $y=n$ in the equation from the statement, we have $f(m)=f(x) f(n)-f(x+n)+1$. Since $f(x+n)=f(x)+n$, it follows that $f(x)=\frac{m}{n}=x$.

## 2 Second Course

## - Statements of exercises covered in class -