45th Mongolian Mathematical Olympiad

Substituting $x=y=0$ in the functional equation, we get $f^2(0)=1$ or $f(0)=\pm 1$.

If $f(0)=1$ then substituting $x=x,y=0$, we get $f(0)=f(x)\cdot f(0)+f(x)-1$ or $f(x)=1$. So in this case $f(x)=1$ is a constant function.

If $f(0)=-1$, then substituting $x=y=2$, we get $f(4)=f^2(2)+f(4)-1$ or $f(2)=\pm 1$.

Now suppose that $f(2)=-1$. Then, substituting $x=y=1$, we get $f(1)=f^2(1)+f(2)-1\iff f^2(1)-f(1)-2=0$. From this quadratic equation we find that $f(1)=-1$ or $f(1)=2$.

If $f(1)=2$ then substituting $x=x$, $y=1$ we get $f(x)=f(x)f(1)+f(x+1)-1\iff f(x+1)=1-f(x)$. From here we get $f(x+2)=1-f(x+1)=f(x)$. Also, another way, substituting $x=x$, $y=2$ then we get $f(2x)=f(x)f(2)+f(x+2)-1=-f(x)+2f(x+1)+1-1=-f(x)+2f(x+1)$. But from above, $f(x+1)=1-f(x)$, so $f(2x)=-f(x)+2(1-f(x))=-f(x)+2-2f(x)=2-3f(x)$. But for every $x\in \mathbb{Q}$ we get that $f(x)=-1$, but we have $x=1$ for $f(1)=2$. Which is a contradiction.

Now if $f(1)=-1$ then substituting $x=x$, $y=1$ we get $$f(x)=f(x)\cdot f(1)+f(x+1)-1\iff f(x+1)=2f(x)+1.(1)$$

On the other hand, we get $$\begin{array}{rl}f(2x)& =f(x)\cdot f(2)+f(x+2)-1=-f(x)+2f(x+1)+1-1=\\ & =-f(x)+2(2f(x)+1)=-f(x)+4f(x)+2=3f(x)+2.\end{array}(2)$$

By using (1) we get that: $$\begin{array}{rl}f(m)& =f(m-1)+1=f(m-2)+2=\cdots =f(2)+m-2=m-1.\end{array}$$ for every $m\in \mathbb{N}$, since $f(2)=1$ (see below).

Now, let's check $f(2)$. If $f(2)=-1$, as above, we get a contradiction. So $f(2)=1$.

Now substituting $x=y=1$, we get $f(1)=f^2(1)+f(2)-1$. So $f(1)=0$ or $f(1)=1$. If $f(1)=1$ then substituting $x=x$, $y=1$ we get $f(x)=f(x)\cdot f(1)+f(x+1)-1$ from this $f(x+1)=1$. Substituting $x=-1$ then $f(0)=1$ which is a contradiction with $f(0)=-1$.

Now we have $f(0)=-1$, $f(1)=0$, $f(2)=1$. For every $x\in \mathbb{Q}$: $f(x)=f(x)\cdot f(1)+f(x+1)-1\iff$ $$\iff f(x+1)=f(x)+1.(3)$$

Then for every $m\in \mathbb{N}$ $$f(m)=f(m-1)+1=f(m-2)+2=\cdots =f(2)+m-2=m-1.$$

Also, $$0=f(1)=f\left(m\cdot \frac{1}{m}\right)=f(m)\cdot f\left(\frac{1}{m}\right)+f\left(m+\frac{1}{m}\right)-1$$ By using (3) $$\begin{array}{rl}f\left(m+\frac{1}{m}\right)& =f\left(m-1+\frac{1}{m}\right)+1=\cdots =f\left(\frac{1}{m}\right)+m\iff \\ \iff 0& =(m-1)f\left(\frac{1}{m}\right)+f\left(\frac{1}{m}\right)+m-1\end{array}$$ So we get $f\left(\frac{1}{m}\right)=\frac{1}{m}-1$.

Now for every $m,n\in \mathbb{N}$, $$\begin{array}{rl}f\left(\frac{n}{m}\right)& =f(n)\cdot f\left(\frac{1}{m}\right)+f\left(n+\frac{1}{m}\right)-1=\\ & =(n-1)\cdot \left(\frac{1}{m}-1\right)+n+\frac{1}{m}-1-1=\frac{n}{m}-1.\end{array}$$

Therefore, the solutions are: for every $x\in \mathbb{Q}$, $f(x)=-1$ or $f(x)=x-1$.