functional equation problem in competition: $ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $

Question

Find all $ f : \mathbb Q \to \mathbb Q $ such that $ f ( 1 ) = 2 $ and $$ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $$ for all $ x , y \in \mathbb Q $.

thank you very much!

Answer 1

Put $y=1$. Than we have $$f(x)=f(x)\cdot f(1)- f(x+1)+1$$ Hence $$f(x)=2f(x)- f(x+1)+1$$ Hence $$f(x+1)=f(x)+1$$. So we have $f(0)=1$. Now we say $x=-y$, we have $$f(-x^2)=f(x)f(-x)$$ When we have $$0=f(-1)=f(-2\cdot \frac{1}{2}) = f(-2) f(\frac{1}{2})- f(-\frac{3}{2}) +1$$ So we have $$0= -1\cdot f(\frac{1}{2}) - f(\frac{1}{2}) +3$$ Hence $$0= -1(2\cdot f(\frac{1}{2})-3) $$ Hence $f(\frac{1}{2})=1$ You need more help?

In general we have for $x\in \mathbb{Z}$ \begin{align*} 0&=f(-1)\\ &=f(-x\cdot \frac{1}{x}) \\ &=f(-x) \cdot f(\frac{1}{x}) - f(-x+\frac{1}{x})+1\\ &= (-x+1)\cdot f( \frac{1}{x}) - f(\frac{1}{x}) +x\\ &= -x \cdot( f(\frac{1}{x}) -1) \end{align*}