pohoatza wrote: Setting $ m=1$ and $ n=t$ , where $ t$ is a positive integer, therefore we obtain $ f(t+1)=f(t)+1$, so $ f(t)=t+1$. Now similar setting $ m=-1$ and $ n=1$, we obtain that $ f(-1)=0$, and by taking $ m=-1$ and $ y=t$ it gives $ f(-t)=-f(t-1)+1=-t+1$. So $ f(z)=z+1$, for each integer $ z$. By plugging $ m=t$ and $ n=\frac{1}{t}$, where $ t$ is an integer., we get $ f(1)=(t+1)f(\frac{1}{t})-f(t+\frac{1}{t})+1$ Furthermore for $ m=1$ and $ y=p+\frac{1}{q}$, $ p,q$ are integers , we have that. $ f(p+1+\frac{1}{q}) = f(p+\frac{1}{q})+1$, hence by induction $ f(p+\frac{1}{q}) = p+f(\frac{1}{q})$. Therefore, $ f(\frac{1}{t}) = \frac{1}{t}+1$ , for each integer $ t$. So, $ f(q)=q+1$, for all rational $ q$.

The solution posted above is not complete; it misses the solution $f(x) = 1$. I believe it is assuming that $f(1) = 2$.

Sorry to revive Let $P(x,y)$ be the assertion of the fonctional equation $f(x+y)=f(x)f(y)-f(xy)+1$ $P(0,0)$ $f(0)=1$ $P(-1,1)$ $f(-1)f(1)=f(-1)$ So we have 2 cases $1-f(1)=1$ $P(x,1)$ $f(x)=1$ for all rational $2-f(-1)=0$ $P(1,1)$ $f(2)=f(1)^2-f(1)+1$ $P(1,2)$ $f(3)=f(1)^3-2f(1)^2+2f(1)$ $P(1,3)$ $f(4)=f(1)^4-3f(1)^3+4f(1)^2-2f(1)+1$ $P(2,2)$ $f(4)=\frac{f(2)^2+1}{2}=\frac{f(1)^4-2f(1)^3+3f(1)^2-2f(1)+2}{2}$ And so solving the equation $f(1)(f(1)-1)^2(f(1)-2)=0$ If $f(1)=2$ $P(x,1)$ $f(x+1)=f(x)+1$ We prove by easy induction that $f(n)=n+1$ and $f(x+n)=f(x)+n$ for all positive integers And from $P(x,-1)$ $f(-x)=1-f(x-1)$ we extend these relations to Z Now let n a integer $P(n,\frac{1}{n})$ $f(n+\frac{1}{n})=f(\frac{1}{n})+n=(n+1)f(\frac{1}{n})-1$ equiavente to $f(\frac{1}{n})=\frac{1}{n}+1$ Now $P(n,\frac{m}{n})$ give $f(\frac{m}{n})=\frac{m}{n}+1$ If $f(1)=0$ $P(x,1)$ $f(x+1)=-f(x)+1$ Hence $f(x+2)=f(x)$ and so $f$is $2-periodique$ and $f(2)=f(0)$ $P(\frac{x}{2},2)$ $f(\frac{x}{2}+2)=f(\frac{x}{2})=f(\frac{x}{2})f(2)-f(x)+1=f(\frac{x}{2})-f(x)+1$ so $ f(x)=1 $ for all $x\in\mathbb{Q}$ contradition with $f(1)=0$ Finally the solutions are : $f(x)=x+1$ for all $x\in\mathbb{Q}$ $f(x)=1$ for all $x\in\mathbb{Q}$

It's exactly K17 just with a little difference in the situation that f(1)=0 , in that case we see that f(2x)=0 but now because we work with rational numbers we get f(x)=0 for all rational x but we know f(0)=1 , a contradiction so this situation doesn't have any answers . the else it's the same K17 .

Let $P(x,y)$ denote the given assertion. $P(0,0)\implies f(0)=1.$ Let $f(1)=c+1,$ where $c$ is any constant. $P(x,1)\implies f(x+1)=cf(x)+1.$ Making some basic substitutions one gets that $c=-1,0,1.$ a. $c=-1$: yields a contradiction which is not hard to see. b. $c=0$: Then $f$ is identically 1. c. $c=1$: Then $P(1,-1)\implies f(-1)=0.$ If $a,b$ are integers then comparing $P(a,1)$ and $P(a,-1)$ we get $f(a)=a+1.$ And $P(a/b,b)\implies 1+a=f(a/b)+bf(a/b)-f(a/b+b)+1.$ But we have that $f(a+b)=f(a)+b,$ so $f(x)=1+x,$ for any rational $x.$ All of the solutions work.

Arefe wrote: It's exactly K17 just with a little difference in the situation that f(1)=0 , in that case we see that f(2x)=0 but now because we work with rational numbers we get f(x)=0 for all rational x but we know f(0)=1 , a contradiction so this situation doesn't have any answers . the else it's the same K17 . K17 is the same except the function is defined on Z.