Rioplatense Mathematical Olympiad

Let $P(x,y)$ denote the assertion $$f(x+f(y+1))+f(xy)=f(x+1)(f(y)+1).$$ If $f$ is a constant $c$, we have $2c=c(c+1)$. This implies $c=0$ or $c=1$. Hence, there are two constant solutions, $f\equiv 0$ and $f\equiv 1$.

$$P(0,y):f(f(y+1))+f(0)=f(1)(f(y)+1),(1)$$ $$P(x-1,0):f(x+f(1)-1)=f(x)(f(0)+1)-f(0).(2)$$ Assume $f(1)=1$. Then (2) becomes $f(x)=f(x)(f(0)+1)-f(0)$, or equivalently, $(f(x)-1)f(0)=0$. Hence $f(0)=0$. Therefore, (1) becomes $$f(f(y+1))=f(y)+1.(3)$$ Setting $y=-1,-2,-3,\dots$ in (3), we conclude inductively that $f(x)=x$ for all $x\le 0$. In addition, $$P(x,-1):f(x)+f(-x)=0,$$ so $f(x)=x$ for all $x\in \mathbb{Z}$, which is clearly a solution. Assume now that $f(1)\ne 1$ and set $c=f(1)-1\ne 0$. If $f(0)+1=0$, then (2) would imply that $f$ is constant. Therefore, it must be $h=f(0)+1\ne 0$. Equation (2) can now be rewritten as $$f(x+c)=hf(x)-h+1.(4)$$ Next we consider $$P(x,1):f(x+f(2))+f(x)=f(x+1)(f(1)+1)(5)$$ $$P(x+c,1):f(x+c+f(2))+f(x+c)=f(x+c+1)(f(1)+1)(6)$$ Using (4), we find that (6) is equivalent to $$h(f(x+f(2))-h+1+hf(x)-h+1=(hf(x+1)-h+1)(f(1)+1).$$ Subtracting $h$ times (5) we get $$2(1-h)=(1-h)(f(1)+1),$$ hence, either $h=1$ or $f(1)+1=2$. The latter is impossible due to our initial assumption that $f(1)\ne 1$. So $h=1$, i.e., $f(0)=0$. Now equation (4) becomes $f(x+c)=f(x)$ and, as $c\ne 0$, $f$ is periodic. Let $M=\max f$ and $-m=\min f$. Note $M,m\ge 0$ because $f(0)=0$. Choose $x,y\in \mathbb{Z}$ such that $f(x+1)=f(y)=M$. Then, the right-hand side of $P(x,y)$ is equal to $M(M+1)$, while the left-hand side is $$f(x+f(y+1))+xy\le M+M=2M.$$ So $M(M+1)\le 2M$, and thus $M=0$ or $M=1$.

Similarly, if we now choose $x,y\in \mathbb{Z}$ such that $f(x+1)=f(y)=-m$, we find $$-m(-m+1)=m^2-m=f(x+f(y+1))+xy\le 2M\le 2,$$ which implies $m\le 2$. So $f(x)\in \{-2,-1,0,1\}$ for all integers $x$. We analyze the possibilities for $f(1)$: $f(1)=-2$: setting $y=1$ in (1) we get $f(f(2))=2$, absurd. $f(1)=0$: in this case, $c=-1$ and $f$ is constant. Absurd. $f(1)=1$: previously excluded. $f(1)=-1$: in this case, $c=-2$, so $f$ is 2-periodic, with $f(x)=f(0)=0$ for all even $x$, and $f(x)=f(1)=-1$ for all odd $x$. However, $P(1,1)$ gives $f(1+f(2))+f(1)=f(2)(f(1)+1)$, so $-2=0$, a contradiction. In conclusion, the only solutions are $f\equiv 0,f\equiv 1$, and $f(x)=x$ for all $x$.