A short solution to a hard problem: Take y=1 and find that $ f(x+1)=f(x)+1$. Then replace y by y+1 and use again the condition given and the fact that $ f(x+1)=f(x)+1$. You will find that $ f(xy+x)=f(xy)+f(x)-1$. Thus $ g(x)=f(x)-1$ is aditive and then it's easy.

You may study this problem in different interesting cases. Solve it for : a) $f:N \rightarrow N$ b) $f:Q \rightarrow Q$ c) $f:R \rightarrow R$ Pierre.

And what about if $f(1) <> 2$

My solution: $f(xy)=f(x)f(y)-f(x+y)+1$ Put $y=1$ to get $f(x+1)=f(x)+1$ .......$*$ An easy induction on $n$ yields two important results: $1)$for $n$=integer , we have $f(n)=n+1$. $2)$ for $n$=integer and $x$=rational we have $f(x+n)=f(x)+n$ Now putting $x=q$ and $y=\frac{p}{q}$ for some integers $p,q$ we have $f\left(\frac{q.p}{q}\right)=f(p)=f(q)f\left(\frac{p}{q}\right)-f\left(q+\frac{p}{q}\right)+1$. Then using the two results we have : $f\left(\frac{p}{q}\right)=1+\frac{p}{q}$. Hence we have $f(r)=1+r$ for all rational $r$.

Easy one: Let $P(x,y)$ be the assertion. $P(x,1)\implies f(x)+f(x+1)=f(x)f(1)+1\implies f(x+1)=f(x)+1$ with $f(1)=2,$ we can easily get $f(x)=x+1\forall x\in\mathbb{Z}$ An easy induction gives $f(x+n)=f(x)+n\forall x,n\in\mathbb{Z}$ $f\left(\frac{p}{q}+q\right)=f\left(\frac{p}{q}\right)+q$ $P\left(\frac{p}{q},q\right)\implies f\left(\frac{p}{q}\cdot q\right)+f\left(\frac{p}{q}+q\right)=f\left(\frac{p}{q}\right)f(p)+1$ $\implies f(p)+f\left(\frac{p}{q}\right)+q=f\left(\frac{p}{q}\right)f(p)+1$ $\implies p+q=qf\left(\frac{p}{q}\right)\implies \boxed{f\left(\frac{p}{q}\right)=\frac{p}{q}+1}$

It can be solved for Q even if the f(1)=2 wasn’t given. Then the answer becomes 1)f(x)=x+1 2)f(x)=1 It gradually reaches f(1)=2 or 0. The 0 part gets ugly.

$P(x,0)\Rightarrow f(0)=1$ $P(x,1)\Rightarrow f(x+1)=f(x)+1\Rightarrow f(x+n)=f(x)+n\Rightarrow f(n)=n+1\forall n\in\mathbb Z$ $P\left(\frac p{q+1},q+1\right)\Rightarrow f\left(\frac pq\right)=\frac pq+1\Rightarrow\boxed{f(x)=x+1}$ which works.

$P(x,0): f(0)=f(x)f(0)-f(x)+1$. $P(x,1): f(x)=2f(x)-f(x+1)+1\implies f(x+1)=f(x)+1$. So $f(x)=x+1\forall x\in \mathbb{Z}$. For rational $\frac{p}{q}$, where $p$ and $q$ are integers, $P\left(\frac{p}{q},q\right): f(p)=f\left(\frac{p}{q}\right)f(q)-f\left(\frac{p}{q}+q\right)+1$. Thus, $p+1=f\left(\frac{p}{q}\right)(q+1)-f\left(\frac{p}{q}\right)-q+1$. So we have $qf\left(\frac{p}{q}\right)=p+q\implies f\left(\frac{p}{q}\right)=\frac{p}{q}+1$. Thus, $\boxed{f(x)=x+1}$, which works.